CHAPTER 1 REACTION KINETIC (PRACTICE MODULE)

ANSWER PRACTICE MODULE CHAPTER 1 SK025

CHAPTER 1.0

REACTION KINETICS

ENERGY PROFILE DIAGRAM

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

FACTOR AFFECTING REACTION RATE
MAXWELL BOLTZMAN DISTIBUTION CURVE

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

ARRHENIUS EQUATION

1.1 Rate of Reaction
Worksheet 1

1 Define reaction rate?
Ans:
Changes in concentrations of reactants (or products) as a function of
time

2. N2O5(g) 2NO2(g) + ½O2(g)

What happens to the concentrations of the all the gas with time?

Ans: [N2O5] decreases, [NO2] and [O2] increase

3. For the reaction A(g) B(g), sketch two curves on the same set of axes that
show:
(a) The formation of product as a function of time
(b) The consumption of reactant as a function of time

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans:

4. Consider the decomposition of N2O5(g)
Ans: 2N2O5(g) → 4NO2(g) + O2(g)

Assume the initial concentration of N2O5(g) is 0.02 mol/L and that none of the

products are present. Make a graph that shows concentrations of N2O5(g), NO2(g),

andO2(g) as a function of time, all on the same set of axes and roughly to scale.

5. Based on the balanced equation:
Ans
2A + B C + 3D

and the following data:

time (s) [C] (M)

0.00 0.218

7.96 0.603

(a) What is the rate of appearance of C?
(b) What is the rate of appearance of D?
(c) What is the rate of disappearance of A?
(d) What is the rate of disappearance of B?

(a) Rate = - d[C] = (0.603– 0.218)
dt 7.96 - 0
= 0.048 M/s

(b)
-1/3 d[D] = d[C]
dt dt
- d[D] = d[C] X 3 = 0.048 X 3
dt dt

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

= 0.145 M/s
(c)

-1/2 d[A] = d[C]
dt dt

- d[A] = d[C] X 2 = 0.048 X 2
dt dt
= - 0.096 M/s

(d)
- d[B] = d[C]
dt dt

= -0.048 M/s

6. Based on the balanced equation:
Ans:
2X + 4Y 3Z + 5Q
7.
and the following data:

time (s) [Y] (M)

0.00 1.358

11.9 0.407

(a) What is the rate of appearance of Y?
(b) What is the rate of appearance of X?
(c) What is the rate of disappearance of Z?
(d) What is the rate of disappearance of Q?

(a)
Rate = - d[Y] = (0.407 – 1.358) = - 0.0799 M/s

dt 11.9 - 0
(b)

-1/2 d[X] = -1/4d[Y]
dt dt

- d[X] =-1/4 d[Y] X 2 = - 1/4 (-0.0799)(2)
dt dt
= -0.03995 M/s

(c) 1/3 d[Z] = -1/4d[Y]
dt dt

d[Z] =-1/4 d[Y] X 4 = - 1/4 (-0.0799)(3)
dt dt
= 0.05992 M/s

(d) 1/5 d[Q] = -1/4d[Y]
dt dt

d[Q] =-1/4 d[Y] X 5 = - 1/4 (-0.0799)(5)
dt dt
= 0.0999 M/s

In the reaction X 2Y, the concentration of X at t = 28.0 s and t = 117.5 s is 2.15
mol dm-3 and 1.08 mol dm-3 respectively. What is the average of the reaction during

this time interval?

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans: (2.15 – 1.08) =
8. (28 – 117.5 )

- 0.012 mol dm-3s-1

Write the rate of reaction in terms of the rate of disappearance of reactant and the

rate of formation of products:

a. H2 (g) + I2 (g) 2 HI(g)

b. 5Br- (aq) + BrO3- (aq) + 6 H+ 3Br2(aq) + 3H2O(l)

c. NO(g) + O3(g) NO2(g) + O2(g)

d. N2(g) + 3H2(g) 2NH3(g)

e. 2C2H6 (g) + 7O2(g) 4CO2(g) + 6H2O(l)

Ans:

9. Consider the reaction between magnesium and hydrochloric acid to form magnesium
Ans chloride and hydrogen gas. When the rate of formation of hydrogen is increasing 0.32
mol dm-3 s-1, what is the rate of decreasing of hydrochloric acid?
10.
Ans Chemical equation: Mg + 2HCl → MgCl2 + H2

-1/2d[HCl] = d[H2]
dt dt
- d[HCI] =-d[H2] X 2 = (0.32)(2)
dt dt
= 0.64 mol dm-3 s-1

Ammonium burns in oxygen to form nitrogen oxide and water.
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

(a) Write the differential rate equation for the reaction.
(b) At particular moment, the rate which oxygen disappear is 0.40 mol dm-3 s-1.

What is the rate of disappearance of ammonia and the rate at which water is
formed?

(a) Rate = -¼ d[NH3] = -1/5 d[O2] = + ¼ d[NO] = + 1/6 [H2O]
dt dt dt dt

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

(b)
-1/4d[NH3] = -1/5 d[O2]
dt dt
- d[NH3] =-1/5 d[O2] X 4 = -(1/5) (- 0.40)(4)
dt dt
= 0.32 mol dm-3 s-1,

(c)
1/6d[H2O] = -1/5 d[O2]
dt dt
- d[H2O] =-1/5 d[O2] X 6 = -(1/5) (- 0.40)(6)
dt dt
= 0.48 mol dm-3 s-1

11. For the reaction : 4NH3(g) + 3O2(g) 2N2(g) + 6H2O (g), The rate of decrease
Ans:
of NH3 is 0.82 mol dm-3 s-1. What is the rate decrease of O2 and the rate of
12.
Ans: formation of N2 and H2O.

= − [ ] = − [ ] = + [ ] = + [ ]

-1/3d[O2] = -1/4 d[NH3]

dt dt

- d[O2] =-1/4 d[NH3] X 3 = -(1/4) (- 0.82)(3)

dt dt

Rate of decrease of O2 = 0.62 mol dm-3 s-1

1/2d[N2] = -1/4 d[NH3]
dt dt

d[N2] =-1/4 d[NH3] X 2 = -(1/4) (- 0.82)(2)
dt dt

Rate of formation of N2=0.41 moldm-3 s-1

1/6d[H2O] = -1/4 d[NH3]
dt dt

- d[H2O] =-1/4 d[NH3] X 6 = -(1/4) (- 0.82)(6)
dt dt

Rate formation of H2O = 1.23 mol dm-3 s-1

Nitrogen pentoxide gas decomposes to nitrogen dioxide gas and oxgen gas

according to the following balanced equation.

2N2N5(g) 4NO2 (g) + O2(g)

(a) Write the differential rate equation for dissociation of N2O5 gas.
(b) If the rate of disappearnace of N2O5 gas is 1.0 X 10-2 mol dm-3 s-1, what is

the rate of formation of NO2 gas under the same condition of temperature

and pressure?

(a) Rate = -1/2 d[N2O5] = ¼ d[NO2] = d[O2]

dt dt dt

(b)

1/4 d[NO2] = -1/2 d[N2O5]

dt dt

= - ½ (-1.0 X 10-2)(4)

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

= 2.0 X 102 mol dm-3 s-1

13. Nitrogen monoxide gas is reduced by hydrogen gas to nitogen gas and steam at
Ans:
9040C.
14.
Ans: 2NO(g) + 2H2(g) N2(g) + 2H2O(g)

15. In one experiment, the rate of formation of nitrogen gas at 904 oC was found to be
Ans:
3.4 X 10-2 mol dm-3 s-1.

(a) Write the differential rate equation for the reaction.

(b) Calculate the rate of disapperance of NO at 9040C.

(c) Predict the rate of disapperance of H2 at 9040C.

(a) Rate = -1/2 d[NO] = - ½ d[H2] = d[N2] = ½ d[H2O]

dt dt dt dt

(b)

Rate = -1/2 d[NO] = d[N2]

dt dt

= (3.4 X 10-2)(2)

= 0.068 mol dm-3 s-1

(c)

Rate = -1/2 d[H2] = d[N2]

dt dt

= (3.4 X 10-2)(2)

= 0.068 mol dm-3 s-1

The rate law below:

H2(g)+ 2NO(g) ⟶ N2O(g)+ H2O(g

has been experimentally determined to be rate = k[NO]2[H2]. What are the orders with
respect to each reactant, and what is the overall order of the reaction?

order in NO = 2
order in H2 = 1
overall order = 3

A series of experiments is performed for the system 2A + 3B + C ⟶ D + 2E
●When the initial concentration of A is doubled, the rate increases by a factor of 4.
●When the initial concentration of B is doubled, the rate is doubled.

●When the initial concentration of C is doubled, there is no effect on rate.

a) What is the order of reaction with respect to each of the reactants?
b) Write an expression for the rate equation.

(a) 2nd order with respect to A
1st order with respect to B

0 order with respect to C
(b) Rate = k[A]2[B]

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

16. CH3OH+CH3CH2OCOCH3⟶CH3OCOCH3+CH3CH2OH

Ans: The rate law for the reaction between methanol and ethyl acetate is, under certain
17. conditions, experimentally determined to be:

Ans: rate=k[CH3OH]
18. What is the order of reaction with respect to methanol and ethyl acetate, and what is
the overall order of reaction?
Ans:
19. order in CH3OH = 1
order in CH3CH2OCOCH3 = 0
overall order = 1

Nitric oxide and chlorine react to form nitrosyl chloride according to the equation

2NO + Cl2 2NOCl

The rate equation for the reaction is:
Rate = k[NO]2[Cl2]

(a) What is the overall order?
(b) What are the unit of k?
(c) What happen to the rate if [Cl2] is doubled?
(d) What happen to the rate if [NO] is doubled?

(a) 3rd order
(b) M-2 s-1
(c) Rate = k[NO]2[2Cl2]

Rate = 2 k[NO]2[Cl2]

Rate is double
(d) Rate = k[2NO]2[Cl2]

Rate = 4k[NO]2[Cl2]

Rate is going up by 4

Consider the reaction A ⟶ B, the rate of reaction is 1.6 X 10-2 M/s, when [A] is 0.35
M. Calculate the rate constant, k, if:

(a) The reaction is first order in A.
(b) The reaction is second order in A.

(a) Rate = k[A]
k = 1.6 X 10-2 M/s

0.35 M
= 4.57 X 10-2 s-1
(b) Rate = k[A]2
k = 1.6 X 10-2 M/s

(0.35 M)2
= 1.31 X 10-1 Ms-1

Consider the following chemical reaction of rusting of iron nail.
4Fe(s) + 3O2 (g) ⟶ 2Fe2O3(S)

If the rate of consumption (disappearance) of O2 (g) is 0.065 M/s, what is the
(a) rate of disappearance of Fe(s)?

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans: (b) rate of formation of Fe2O3(S)
20.
(a)
Rate = -1/4 d[Fe] = -1/3d[O2]
dt dt
= -1/3(-0.065) (4)
= 0.0867 M/s

(b)
Rate = -1/2 d[Fe2O3] = 1/3d[O2]
dt dt
= -1/3(-0.065) (2)
= 0.0433 M/s

The decomposition of N2O5 proceeds according to the following equation:
2N2O5(g) → 4NO2(g) + O2(g)

If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is
4.2×10–7 M/s, what is the rate of appearance of (a) NO2, (b) O2?

Ans: (a)

Rate = 1/4 d[NO2] = - 1/2d[N2O5]
dt dt
= -1/2(4.2×10–7) (4)
= 8.4 ×10–7 M/s

(b)
Rate = d[O2] = - 1/2d[N2O5]
dt dt
= -1/2(4.2×10–7)

= 2.1 ×10–7 M/s

21. The following data were measured for the reaction of nitric oxide with hydrogen:

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans: (a) Determine the rate law for this reaction.
22. (b) Calculate the rate constant.
(c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M

(a)
rate = k[NO]x[H2] y
Order reaction of H2
Rate 2 = k[NO]x[H2]y
Rate 1 = k[NO]x[H2] y
2.46 X 10-3 = k (0.1)x(0.020)y
1.23 X 10-3 k (0.1) x (0.010) y
2 = 2y
y= 1
1st order reaction in H2.

Order reaction of NO
Rate 3 = k[NO]x[H2]y
Rate 1 = k[NO]x[H2] y
4.92 X 10-3 = k (0.2)x(0.1)y
1.23 X 10-3 k (0.1) x (0.1) y
4 = 2x
x= 2
2nd order reaction in NO.
rate = k[NO]2[H2].

(b)
From exp 1:
1.23 X 10-3 M/s = k (0.1) 2 (0.1)
k= 1.23 M–2 s–1

(c)
rate = k[NO]2[H2].

= (1.23) (0.05)2 (0.15)
rate = 4.6125 ×10–4 M/s

Consider a hypothetical reaction:

Ans: a) Determine the rate law expression for this reaction.
b) What is the order of reaction with respect to A? to B?
c) What is the overall reaction order?

(a)
rate = k[A]x[B] y

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Order reaction of B
Rate 2 = k[A]x[B]y
Rate 1 = k[A]x[B] y
4 = k (0.1)x(0.20)y
4 k (0.1) x (0.10) y
1 = 2y

y= 0

0 order reaction in B.

Order reaction of A
Rate 3 = k[A]x[B]y
Rate 1 = k[A]x[B]y
16 = k (0.2)x(0.1)y

4 k (0.1) x (0.1) y
4 = 2x

x= 2
2nd order reaction in A.
rate = k[A]2[B].

(b) 2nd order with respect to A

0 order with respect to B
(c) 2nd order

23. Acetaldehyde decomposes when heated to yield methane and carbon monoxide

according to the equation:

CH3CHO(g)⟶CH4(g)+CO(g)

Determine the rate law and the rate constant for the reaction from the following

experimental data:

Exp [CH3CHO] mol/L Rate (mol/L s-1)
1 1.75 × 10−3 2.06 × 10−11

2 3.50 × 10−3 8.24 × 10−11

3 7.00 × 10−3 3.30 × 10−10

Ans: Rate = k[CH3CHO]x

Rate 2 = k[CH3CHO]x
Rate 1 = k[CH3CHO]x
8.24 x 10-11 = k (3.50 x 10-3)x
2.06 x 10-11 k (1.75 x 10-3) x
4= 2x
x=2
2nd order reaction in CH3CHO
Rate = k[CH3CHO]2

From exp 1:
2.06 x 10-11 M = k(1.75 x 10-3 M/s)2
k = 6.73 X 10-6 L/mol/s @ M-1 s-1

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

24. Using the initial rates method and the experimental data, determine the rate law
Ans:
and the value of the rate constant for this reaction:
25.
2NO(g)+Cl2(g)⟶2NOCl(g)

Exp [NO] mol/L [Cl2] mol/L Rate
(mol/L s-1)

1 0.10 0.10 0.00300

2 0.10 0.15 0.00450

3 0.15 0.10 0.00675

rate = k[NO]x[Cl2] y

Order reaction of NO
Rate 3 = k[NO]x[Cl2] y
Rate 1 = k[NO]x[Cl2] y
0.00675= k (0.15)x(0.10)y
0.00300 k (0.10) x (0.10) y
2.25 = 1.5x

x= 2
2nd order reaction in NO.

Order reaction of Cl2
Rate 2 = k[NO]x[Cl2] y
Rate 1 = k[NO]x[Cl2] y
0.00450 = k (0.10)x(0.15)y
0.00300 k (0.10) x (0.10) y
1.5 = 1.5y

y= 1
1st order reaction in Cl2.
Rate = k[NO]2[Cl2]

(b)

From exp 1:
0.00300 M/s = k (0.1)2 (0.1)
k = 3.0 mol−2 L2 s−1

Use the provided rate data to derive the rate law for the reaction whose equation is:
OCl−(aq)+I−(aq)⟶OI−(aq)+Cl−(aq)

Exp [OCl-] mol/L [l-] mol/L Rate
(mol/L s-1)
1 0.0040 0.0020 0.00184
2 0.0020 0.0040 0.00092
3 0.0020 0.0020 0.00046

Ans: rate = k[OCl-]x[l-]y
Order reaction of OCl-
Rate 1 = k[OCl-]x[l-]y
Rate 3 = k[OCl-]x[l-]y
0.00184= k (0.004)x(0.002)y
0.00046 k (0.002) x (0.002) y

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

4 = 2x

x= 2
2nd order reaction in OCl-

Order reaction of I-
Rate 2 = k[OCl-]x[l-]y
Rate 3 = k[OCl-]x[l-]y
0.00092 = k (0.002)x(0.004)y
0.00046 k (0.002) x (0.002) y
2 = 2y

y= 1
1st order reaction in I-.
Rate = k[OCl-]2[l-]

26. Consider the reaction: P4 + 6H2 → 4PH3. A rate study of this reaction was
Ans:
conducted at 298 K. The data that were obtained are shown in the table.

Exp [P4] mol/L [H2] mol/L Rate
(mol/L s-1)

1 0.0110 0.0075 3.20x10-4

2 0.0110 0.0150 6.40x10-4

3 0.0220 0.0150 6.39x10-4

(a) What is the order with respect to P4 and H2?
(b) Write the rate law for this reaction.
(c) Determine the value and units of the rate constant, k.

(a)
rate = k[P4]x[H2]y

Order reaction of P4
Rate 2 = k[P4]x[H2]y
Rate 3 = k[P4]x[H2]y
6.40 X 10-4= k (0.011)x(0.015)y
6.39 X 10-4 k (0.022) x (0.015) y1
1 = 0.5x

x= 0

0 order reaction in P4

Order reaction of H2
Rate 2 = k[P4]x[H2]y
Rate 1 = k[P4]x[H2]y
6.40 X 10-4= k (0.011)x(0.015)y
3.20 X 10-4 k (0.011) x (0.0075) y
2 = 2y

y= 1
1st order reaction in H2

(b) Rate=k[H2]

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

(c)
From exp 1:
3.20x10-4 M/s = k (0.0075 M)
k = 0.0427 s-1

27. With the given information, determine the rate law, the rate constant, and the

overall reaction order.

2Mg + O2 2MgO

Exp [Mg] mol/L [O2] mol/L Initial Rate
(mol/L s-1)
1 0.10 0.10 2.0x10-3
2 0.20 0.10 4.0x10-3
3 0.10 0.20 8.0x10-3

Ans: rate = k[Mg]x[O2]y
Order reaction of Mg
Rate 2 = k[Mg]x[O2]y
Rate 1 = k[Mg]x[O2]y

4.0 X 10-3= k (0.20)x(0.1)y
2.0 X 10-3 k (0.10) x (0.1) y1
2 = 2x
x=1
1st order reaction in Mg

Order reaction of O2
Rate 3 = k[Mg]x[O2]y
Rate 1 = k[Mg]x[O2]y
8.00 X 10-3= k (0.10)x(0.20)y
3.20 X 10-4 k (0.10) x (0.10) y
4 = 2y
y= 2
2nd order reaction in O2
rate law = k[Mg]1[O2]2

From exp 1:
2.0x10-3 M/s = k (0.10) (0.10)2
k= 2.0 M-2s-1

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

28. The reaction between C and D is represented as follows:
Ans: C+D E

Exp [C] mol/L [D] mol/L Time Interval Change in conc.
(minutes) of C (M)
1 0.10 1.0 30 2.5x10-3
2 0.10 2.0 30 1.0x10-2
3 0.05 1.0 120 5.0x10-3

(a) Calculate the rate of reaction of each experiment.
(b) Write the rate law.
(c) State the effect on the concentration rate if the concentration of D is doubled

but the concentration of C remains.

(a) Exp 1:
2.5 x 10-3 = 8.3 X 10-5 M/min
30
Exp 2:
1.0 x 10-2 = 3.3 X 10-4 M/min
30

Exp 2:
5.0 x 10-3 = 4.2 X 10-5 M/min

120

(b)
rate = k[C]x[D]y

Order reaction of D
Rate 3 = k[C]x[D]y
Rate 1 = k[C]x[D]y

5.0x10-3= k (0.10)x(1.0)y
2.5x10-3 k (0.05) x (1.0) y
2 = 2x

x=1
1st order reaction in C
rate = k[C]x[D]y

Order reaction of D
Rate 1 = k[C]x[D]y
Rate 2 = k[C]x[D]y

2.5x10-3= k (0.10)x(1.0)y
1.0x10-2 k (0.10) x (2.0) y
0.25 = 0.5x

x=2
2nd order reaction in D

Rate = k [C][D]2

(c)
Rate = k [C][2D]2
Rate = 4 k[C][D]2

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Reaction increases 4 times.

29. At 4000C, an organic compound A decomposes to form another organic compound.

Ans: The initial rate data are given in the following table.

30. Exp [A] mol/L Initial Rate (mol/L s-1)
Ans:
1 1.8 × 10−2 9.0 × 10−4
31.
Ans: 2 3.0× 10−2 1.5 × 10−3

32. (a) Calculate the rate constant.
(b) Calculate the rate of decomposition of A when the [A] is 6.8 X 10-3 mol/L

(a) rate = k[A]x

Order reaction of A
Rate 2 = k[A]x
Rate 1 = k[A]x

1.5 × 10−3= k (3.0× 10−2)x
9.0 × 10−4 k (1.8 × 10−2) x
1.667 = 1.667x

x=1
1st order reaction in A

Rate = k[A]

From exp 1:
9.0 × 10−4M/s = k (1.8 × 10−2 M)
k=0.05 s-1

(b) Rate = k[A]
Rate = 0.05 s-1 (6.8 X 10-3 mol/L)
3.40 X 10-4 mol/L s-1

The reaction 2A B is first order in A with a rate constant of 2.8 x 10–2 s–1 at

80oC. How long will it take for A to decrease from 0.88 M to 0.14 M?

ln[A0]/[At]=kt
ln (0.88/0.14) = 2.8 x 10–2 t
t = 65.6 s

2N2O5(g) → 2N2O4(g) + O2(g)
The rate law is: Rate = k[N2O5]. At 45oC, the rate constant for the reaction is
6.22 x 10–4 s–1. If the initial concentration of N2O5 in the solution is 0.100 M, how
many minutes will it take for the concentration to drop to 0.0100 M?

ln[N2O5]0/[ N2O5]t=kt
ln (0.100 /0.0100) = 6.22 x 10–4 t
t=61.7 min

Iodine atoms combine to form molecular iodine in the gas phase

I(g) + I(g) I2(g)

The rate constant of the reaction is 7.0 x 109 M-1s-1 at 23oC. If the initial

concentration of I was 0.086 M, calculate the concentration after 2.0 min.

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans: 1/[I]t – 1/[I]0 = kt
33. 1/[I]t = (7.0 x 109) (2.0 x 60) + 1/ (0.086)
[I]t = 1.2 x 10–12 M
Ans:
34. At 25oC, hydrogen iodide breaks down slowly to hydrogen and iodine: rate = k[HI]2.
The rate constant at 25oC is 2.4 x 10–21 L mol-1s-1. If 0.01 mol of HI(g) is placed in a
Ans: 1.0–L container, how long (in years) will it take for the concentration of HI to reach
35. 0.009 mol L-1?

Ans: 1/ [ HI]t – 1/[ HI]0 = kt
36. 1/ (0.009) – 1/ (0.01) = 2.4 x 10–21 t
t= 4.6296 x 1021
Ans: = 1.5 x 1014 years

N2O5 decomposes into N2O4 and O2.
2N2O5(g) → 2N2O4(g) + O2(g)

At 45oC, k = 6.22 x 10–4 s–1.
If the initial concentration of N2O5 in a carbon tetrachloride solution at 45oC is 0.5 M,
what will the concentration be after exactly one hour?

ln[N2O5]t = ln[N2O5]0-kt
= ln (0.5) – (6.22 x 10–4)(1 X 60 X 60)
= 0.053 M

Nitrosyl chloride, NOCl, decomposes slowly to NO and Cl2.
2NOCl(g) → 2NO(g) + Cl2(g)

The rate law shows that the rate is second order in NOCl. Rate = k[NOCl]2
The rate constant equals 0.020 L mol–1s–1 at certain temperature. Determine how
many minutes it would take for the NOCl concentration to drop from 0.040 M to
0.010 M.

1/[ NOCl]t – 1/[ NOCl]0 = kt
1/ (0.010) – 1/(0.040) = 0.020 t
t= 3750 s
= 62.5 min

In a first order decomposition reaction, 50% of a compound decomposes in 10.5
min.
(a) What is the rate constant of the reaction?
(b) How long does it take for 75.0% of the compound to decompose?

a) t ½ = ln 2/k
10.5 = ln 2 / k
k = 6.60 x 10-2 min–1

b) kt = ln ([A]0/[A]t)
6.60 x 10-2 (t) = ln (100/25)
t = 21.0 min

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

37. Use a value of k = 7.30 x 10–4 s–1 for the first–order decomposition of H2O2(aq) to
Ans: determine the percentage H2O2 has decomposed in the first 500.0 s after the
reaction begins.

ln[H2O2]t = ln[H2O2]0-kt
= ln (100) – (7.30 x 10–4) (500)
= 69.4

% decomposition = 100 - 69.4 = 30.6 %

38. Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as

follows:
C2H5I(g) → C2H4(g) + HI(g)

From the following data determine the order of the reaction and the rate constant.

TIME (min) [ C2H5I] (M)
0 0.36
15 0.30
30 0.25
48 0.19
75 0.13

Ans:

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

39. = 1.3 x 10–2 min–1
Ans:
In decomposition reaction B → products, the following data are obtained:
t = 0 s, [B] = 0.88 M; t = 25 s, 0.74 M; t = 50s, 0.62 M; t = 75 s, 0.52 M; t = 100 s,
0.44 M; t = 150 s, 0.31 M; t = 200 s, 0.22 M; t = 250 s, 0.16 M.
What is the order of this reaction and its rate constant k?

40. k = 6.93 x 10–3 s–1
Ans:
What is the half–life (in min) of N2O5 if it decomposes with a rate constant of
41. 5.7 x 10–4 s–1?

a) t ½ = ln 2/k
= ln 2 / 5.7 x 10–4
= 1216 s
= 20.3 min

Cyclopropane(C3H6) is the smallest cyclic hydrocarbon because its 60oC bond
angles allow only poor orbital overlap, it bonds are weak. As a result, it is thermally
unstable and rearrange to propene:

The rate constant is 9.2 s–1.
(a) What is the half–life of the reaction?
(b) How long does it take for the concentration to reach one-quarter of the

initial value?

20

ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans: (a) t ½ = ln 2/k
= ln 2 / 9.2
42. = 0.075 s
Ans:
(b) ln[C3H6]0/ [ C3H6]t=kt
43. ln (100/25) = 9.2 t
t = 0.15 s

The thermal decomposition of N2O5 obeys first–order kinetics. At 45oC, a plot of ln
[N2O5] versus t give slope of – 6.18 x 10–4 min–1. What is the half–life of the
reaction?

The reaction is first–order overall.

t ½ = ln 2/k
= ln 2 / 6.18 x 10–4
= 1.12 x 103 min

During a study of ammonia production, an industrial chemist discovers that the
compound decomposes to N2 and H2 in a first–order process. He collects the

following data:

Time (s) 0 1.0 2.0

[NH3] 4.000 3.986 3.974
(mol/L)

Ans: (a) Use graphical method to determine the rate constant
(b) What is the half–life for ammonia decomposition?

(a)

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

The reaction is first–order. k= 3 x 10–3 s–1
(b)

44. The rate constant of the reaction 2A → B is 51 M–1 min–1 at 24oC.
Ans: (a) Starting with [A] = 0.0092 M, how long will it take for [A] = 3.7 x 10–3 M?
(b) Calculate the half–life of the reaction.
45.
Ans: (a) The reaction is second–order overall.
1/[ A]t – 1/[A]0 = kt
1/(3.7 x 10–3 ) – 1/(0.0092) = 51t
t = 3.2 min

(b) t1/2 = 1/ k[A]0
= 1/ ((51) (0.0092))
= 2.1 min

The reaction 2HI(g) → H2(g) + I2(g) has the rate law = k[HI]2, with k = 0.079 L mol–
1s–1 at 508oC. What is the half–life for this reaction at this temperature when the
initial HI concentration is 0.050 M?

t1/2 = 1/ k[HI]0
= 1/ ((0.079) (0.050))
= 253.2 s

46. Mercury (II) chloride reacts with oxalate ion according to the following equation:

2HgCl2(aq) + C2O42-(aq)→ 2Cl-(aq) + 2 CO2(g) + Hg2Cl2(s)

The results of the experiment are shown in TABLE 1.

TABLE 1

Experiment [HgCl2] [C2O42-] Initial rate
(moldm-3) (moldm-3) (moldm-3 h-1)
I
II 0.632 0.049 1.0x 10-3
III 1.0x 10-3
0.948 0.049 4.0x 10-3

0.632 0.098

i. Determine the order of reaction with respect to HgCl2 and C2O42-.

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

ii. Calculate the value of the rate constant.
iii. If the initial concentration of C2O42- is 2.8 M, what is the concentration of
C2O42after 3.25 hours.
iv.
Ans: i. rate = k[HgCl2]x[C2O42-] y
Order reaction of HgCl2
Rate 2 = k[HgCl2]x[C2O42-]y
Rate 1 = k[HgCl2]x[C2O42-]y
1.0x 10-3= k (0.948) x (0.049) y
1.0x 10-3 k (0.632) x (0.049) y
1 = 1.5x
x=0
Zero order reaction in HgCl2

rate = k[HgCl2]x[C2O42-] y

Order reaction of HgCl2
Rate 3 = k[HgCl2]x[C2O42-]y
Rate 1 = k[HgCl2]x[C2O42-]y
4.0x 10-3= k (0.632) x (0.098) y
1.0x 10-3 k (0.632) x (0.049) y
4 = 2y

y=2
2nd order reaction in C2O42-

ii.

From exp 1:
1.0x 10-3 moldm-3 h-1= k (0.049 M)2
k = 0.417 moldm-3 h-1

iii.
1/[C2O42-]t = kt + 1/[C2O42-]0

= (0.417) (3.25) + 1/2.8
[C2O42-]t = 0.585 M

47. Define order of reaction and activation energy. The following reaction is first order with

respect to NO2 and zero order with respect to O2.
2N02 + 1/2 O2 → N2O5

In an experiment, the rate constant, k, was determined at different temperatures and

the data were tabulated in TABLE 1.

TABLE 1 k (s-1)
Temperature (K) 1.8 x 10-4
2.7 x 10-3
477 3.0 x 10-2
523 2.6 x 10-1
577
667

Write the rate of law of this reaction and determine graphically the activation energy,
Ea, (in kJ mol-1) for the reaction.

23

ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans Order of reaction is an exponent in a rate law with respect to the
corresponding reactant.
Activation energy is the minimum kinetic energy that has must be
processed by the reactants in order to result in an effective collision.
Rate = k [NO₂]
Ea = (12.167 x 10³ x 8.314) J mol-1
= 1.01 x 10² kJmol-1

1.2 Collision Theory
Worksheet 2

1. Explain collision theory.

Ans: Collision theory is a way to explain why different reactions occur at
different rates. It states:

i. Molecules of reactants must collide in order to form products.
ii. Products are formed only when effective collisions occur between

molecules.

2. What are the two things that must take place in order for a reaction to take place
between molecules or atom?

Ans: i. The colliding molecules must have a total kinetic energy equal to
or greater than the activation energy, Ea.

ii. Collisions occur at the correct orientation.

3. Describe how the activation energy of a reaction affects the overall rate of the
chemical reaction.

Ans: The higher the Ea, the more energy required during the collision for the
reaction occur.

4. The diagram shows the reaction occur.

The following collisions do NOT react. State why.
(a)
(b)

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

(c)

Ans: (a) Not correct orientation
5. (b) Not enough energy
(c) Not correct orientation

Draw Label an Endothermic and Exothermic graph of A + B C + D. Make

sure you include the Products, Reactants, Activation Energy, Activated Complex,
and the Heat of the Reaction (ΔH).

Ans: Endothermic Exothermic

6. A key reaction in the upper atmosphere is
Ans: O3(g) + O(g) → 2O2(g)

The Ea(fwd) is 19 kJ, and ∆Hrxn for the reaction is – 392 kJ. Draw a reaction energy

diagram and calculate Ea(rev).

7. For the reaction A2 + B2 → 2AB, Ea(fwd) = 125 kJ/mol and Ea(rev) = 85 kJ/mol.
Assuming the reaction occurs in one step,
(a) Draw a reaction energy diagram.
(b) Calculate ∆Hrxn.

25

ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans:

8. (a)
Ans: (b) 125 kJ/mol – 85 kJ/mol = 40 kJ/mol

The reaction A → B has a reaction enthalpy of – 64 kJ/mol and an activation
energy of 22 kJ/mol.
(a) What is the activation energy of the B → A reaction.
(b) Draw the energy profile for the reaction.

(a) 64 kJ/mol + 22 kJ/mol = 86 kJ/mol

9. For the reversible reaction A + B → C + D, the enthalpy change of the forward
Ans: reaction is +21 kJ/mol. The activation energy of the forward
reaction is 84 kJ/mol.
(a) What is the activation energy of the reverse

reaction?
(b) Sketch the reaction profile of this reaction.

a) 63 kJ/mol

26

ANSWER PRACTICE MODULE CHAPTER 1 SK025

10. i. Which of the letters a–f in the
Ans: diagram represents the potential
energy of the products? ___

ii. Which letter indicates the
potential energy of the activated
complex? ______

iii. Which letter indicates the
potential energy of the reactants?
___

iv. Which letter indicates the
activation energy? ______

v. Which letter indicates the heat
of reaction? ______

vi. Is the reaction exothermic or endothermic? ________
vii. Which letter indicates the activation energy of the reverse reaction? ___
viii. Which letter indicates the heat of reaction of the reverse reaction? ___
ix. Is the reverse reaction exothermic or endothermic? _______

i. e
ii. c
iii. a
iv. b
v. f
vi. endothermic
vii. d
viii. f
ix. exothermic

1.3 Factor affecting reaction rate
Worksheet 3

1. Complete the following table by indicating which factor would have the
greatest impact on the rate of reaction. Choose from concentration,
temperature, surface area or catalyst.

SCENARIO FACTOR
Raw carrots are cut into thin slices for cooking.
Protein is broken down in the stomach by the enzyme
pepsin.
A Woolly Mammoth is found, perfectly preserved, near the
Arctic circle.
More bubbles appear when a concentrated solution of
hydrochloric acid is added to a magnesium strip than
when a dilute solution of the acid is added.

27

ANSWER PRACTICE MODULE CHAPTER 1 SK025

Exhaust from a car engine passes through a catalytic
converter changing most of the poisonous nitrogen oxides
into nitrogen gas and oxygen gas.
A dust explosion occurs in a saw mill.
Blowing air on a campfire to help get it going.

Ans: SCENARIO FACTOR

Raw carrots are cut into thin slices for cooking. Surface area

Protein is broken down in the stomach by the Catalyst

enzyme pepsin.

A Woolly Mammoth is found, perfectly preserved, Temperature

near the Arctic circle.

More bubbles appear when a concentrated solution Concentration

of hydrochloric acid is added to a magnesium strip

than when a dilute solution of the acid is added.

Exhaust from a car engine passes through a catalytic Catalyst

converter changing most of the poisonous nitrogen

oxides into nitrogen gas and oxygen gas.

A dust explosion occurs in a saw mill. Concentration

Blowing air on a campfire to help get it going. concentration

2. A freshly exposed surface of metallic sodium tarnishes almost instantly if exposed to air
and moisture, while iron will slowly turn to rust under the same conditions. In these two
situations, the__________________ refers to how quickly or slowly reactants turn into
products

Ans: Rate of reaction

3. ___________________ refers to how much solute is dissolved in a solution. If there is
a greater concentration of reactant particles present, there is a greater chance that
__________________ among them will occur. More collisions mean a higher rate of
reaction.

Ans Concentration, collision

4. Grains of sugar have a greater ___________________ than a solid cube of sugar of the
same mass, and therefore will dissolve quicker in water.

Ans: Surface area

5. Chewing antacid tablets is better than swallowing them whole. Why?

Ans: Chewing increases the surface area so the tablets dissolve faster.

28

ANSWER PRACTICE MODULE CHAPTER 1 SK025

6. Explain the effect of temperature on reaction rate using Maxwell-Boltzman Distibution
Curve

Ans: Maxwell-Boltzmann distribution shows the kinetic energy distributions for a reaction
mixture at two different temperatures.

Shaded area

• T 2 >T 1: At T 2, molecules that have kinetic energy equal or greater than
activation energy is more compared to T1.

• The area under the curve equal to the total number of molecules. Number
of molecules below both curves are the same.

• The shaded area represents number of molecules possessing kinetic
energy, KE  activation energy, Ea

• At higher temperature (T2), the peak of the curve moves to the right
the curves flatten, more molecules with higher kinetic energy, KE (larger
shaded area). Particles absorb energy and become more energetic
More molecules possess kinetic energy equal to or higher than Ea, higher
frequency of effective collision. So, reaction rate at T2 increases.

7. Draw energy profile diagram showing the difference for an endothermic
reaction in the presence of catalyst and without catalyst.

Ans:

8. Does a catalyst increase reaction rate by the same means as a rise in temperature
does? Explain.

Ans: No. A catalysts increases the reaction rate by providing an alternative
pathway with a lower activation energy.

9. The following gas−phase decomposition reaction is the first order:
C2H5Cl → C2H4 + HCl

In a table of kinetics data, we find the following values listed for this reaction:

29

ANSWER PRACTICE MODULE CHAPTER 1 SK025

A = 1.58 x 1013 s−1, Ea = 237 kJ/mol.
(a) Calculate the value of the specific rate constant at room temperature 25oC.
(b) Calculate the value of the specific rate constant at room temperature 275oC.

Ans: (a) k = Ae- Ea/RT
= 1.58 x 10-3 X e – ((237000)/ (8.314) (298.15))
= 4.52 x 10−29 s−1

(b) k = Ae- Ea/RT
= 1.58 x 10-3 X e – ((237000)/(8.314)(548.15))
= 4.05 x 10−10 s−1

10. Consider the decomposition of NO2 into NO and O2:
2NO2(g) → 2NO(g) + O2(g)

The following data were collected for the reaction.

Rate constant, k Temperature
(L mol–1 s–1 ) (oC)
400
7.8 410
10 420
430
14 440

18

24

Determine graphically the Ea for the reaction.

Ans: Plot graph: ln k versus 1/T
Slope = – Ea/R = – 1.44 x 104 K

Ea = 1.2 x 102 kJ/mol

OBJECTIVE QUESTION

Worksheet 4

1. What is rate of reaction?
A. How fast a reaction is
B. How big a reaction is
C. How loud a reaction is
D. How much gas a reaction produces

2. Which of the following statements is correct?
A. The rate of a reaction decreases with passage of time as the concentration
of reactants decreases.
B. The rate of a reaction is same at any time during the reaction.

30

ANSWER PRACTICE MODULE CHAPTER 1 SK025

C. The rate of a reaction is independent of temperature change.
D. The rate of a reaction decreases with increase in concentration of reactant(s).

3. Which of the following correctly describes processes that happen during reactions?
A. Bonds are broken in reactants, which is an exothermic process that takes in energy
B. Bonds are broken in reactants, which is an endothermic process that gives out
energy
C. Bonds are made in products, which is an endothermic process that takes in energy
D. Bonds are made in products, which is an exothermic process that gives out
energy

4. For a reaction: 1/2A → 2B, rate of disappearance of 'A' is related to the rate of
appearance of 'B' by the expression:

A

B

C

D

5. In the formation of S02 by contact process; 2SO2 + O2 → 2SO3, the rate of reaction was

measured as = 2.5 × 10-4 mol L-1s-1. The rate of formation of S03 will be

A. -5.0 × 10-4 mol L-1s-1

B. -1.25 × 10-4 mol L-1s-1

C. 3.75 × 10-4 mol L-1s-1

D. 5.00 × 10-4 mol L-1s-1

6. For a reaction 2A + B 2C, with the rate equation: Rate = k[A]2[B]
A. the order with respect to A is 2 and the order overall is 2.
B. the order with respect to A is 2 and the order overall is 3.
C. the order with respect to B is 2 and the order overall is 2.
D. the order with respect to B is 2 and the order overall is 3.

31

ANSWER PRACTICE MODULE CHAPTER 1 SK025

7. A reaction was found to be zero order in A. Increasing the concentration of A by a factor
of 3 will cause the reaction rate to __________.
A. remain constant
B. increase by a factor of 27
C. increase by a factor of 9
D. triple

8. The rate law of the overall reaction is k[A][B]0 Which of the following will not increase the
rate of the reaction?
A. increasing the concentration of reactant, A
B. increasing the concentration of reactant B
C. increasing the temperature of the reaction
D. adding a catalyst for the reaction

9. For the reaction A+ 3B 2C, how does the rate of disappearance of B compare to

the rate of production of C?

A. the rate of disappearance of B is 1/2 the rate of appearance of C

B. the rate of disappearance of B is 3/2 the rate of appearance of C

C. the rate of disappearance of B is 2/3 the rate of appearance of C

D. the rate of disappearance of B is 1/3 the rate of appearance of C

10. Given: A + 3B 2C + D

This reaction is first order with respect to reactant A and second order with respect to

reactant B. If the concentration of A is doubled and the concentration of B is halved, the

rate of the reaction would _____ by a factor of _____.

A. increase, 2
B. decrease, 2
C. increase, 4
D. decrease, 4

11. rate = k[A]2
For the reaction for which rate law is given above, a plot of which of the following is a
straight line?

A. versus time.
[ ]

B. [A] versus 1



C. [A] versus time

D. ln [A] versus 1


32

ANSWER PRACTICE MODULE CHAPTER 1 SK025

12. The decomposition of carbon disulfide, CS2, to carbon monosulfide, CS, and sulfur is
first order with k = 2.8 x 10-7 s-1 at 1000oC.
CS2 CS + S
What is the half-life of this reaction at 1000oC?

A. 5.0 x 107 s
B. 4.7 x 10-6 s
C. 3.8 x 105 s
D. 2.5 x 106 s

13. The half-life for a first-order reaction is 32 s. What was the original concentration if, after
2.0 minutes, the reactant concentration is 0.062 M?

A. 0.834 M
B. 0.069 M
C. 0.091 M
D. 0.075 M

14. Given the following data for this reaction:

NH4+(aq) + NO2-(aq) N2(g) + 2H2O(l)

Exp [NH4+] [NO2-] Rate
1 0.010 M 0.020 M 0.020 M/s
2 0.015 M 0.020 M 0.030 M/s
3 0.010 M 0.010 M 0.005 M/s

The rate law for the reaction is:

A. Rate = k[NH4+][NO2-]
B. Rate = k[NH4+]2[NO2-]2
C. Rate = k[NH4+]2[NO2-]
D. Rate = k[NH4+][NO2-]2

15. A reaction is first order with respect to [X] and second order with respect to [Y]. When [X]
is 0.20 M and [Y] = 0.20 M the rate is 8.00 × 10-3 M/min. The value of the rate constant,
including correct units, is

A. 1.00 M min-1
B. 1.00 M-2 min-1
C. 2.00 M-1 min-1
D. 2.00 M-2 min-1

16. A substance B decomposes by a first-order reaction starting initially with [B]=2.00M and
after 200 min, [B] becomes 0.15M. For this reaction t1/2 is
A. 53.52 min
B. 50.49 min

33

ANSWER PRACTICE MODULE CHAPTER 1 SK025

C. 48.45 min
D. 46.45 min
17. What is the term that used to refer the number of collisions per unit volume of the
reaction mixture?
A. Collision force
B. Collision frequency
C. Collision energy
D. Collision time period
18. As the temperature of a reaction is increased, the rate of the reaction increases because
the __________.
A. reactant molecules collide less frequently
B. reactant molecules collide more frequently and with greater energy per
collision
C. activation energy is lowered
D. reactant molecules collide less frequently and with greater energy per collision

19. Crushing a solid into a powder will increase reaction rate because:
A. the particles will collide with more energy
B. the orientation of colliding particles will be improved
C. the activation energy barrier will be lowered
D. the powdered form has more surface area

20. The decomposition of hydrogen peroxide is catalysed by adding a small amount of
manganese (IV) oxide.

34

ANSWER PRACTICE MODULE CHAPTER 1 SK025

21. The minimum amount of energy required for a reaction to proceed.
A. activation energy
B. exothermic energy
C. endothermic energy
D. catalyst

22. For the reaction profile diagram shown, which one of the following statements is correct?

A. A represents the energy of the products, and B is the activation energy.
B. A represents the energy of the reactants and C represents the energy of the

products.
C. A represents the energy of the reactants and B represents the energy of the products.
D. B represents the energy of the products and C represents the energy of the reactants.
23. The rate of a chemical reaction normally

A. increases as temperature decreases.
B. decreases when a catalyst is added.
C. increases as reactant concentration increases.
D. decreases as reactant concentration increases.
24. A lump of ignited charcoal which is glowing in air burns more vigorously when lowered
into a bottle of pure oxygen. This is due to an increase in

A. surface area
B. concentration
C. temperature
D. volume

35

ANSWER PRACTICE MODULE CHAPTER 1 SK025

25. It is generally believed that catalysts increase reaction rates by:
A. removing the activation energy barrier
B. providing an alternate activation energy barrier that is lower than the original
barrier
C. lowering the activation energy barrier
D. giving the reacting particles more energy, thus there will be more successful collisions

26. Which reaction is fastest?
A. 1g limestone powder, 100 cm3 of 1 M acid and 30oC
B. 1g limestone solid,100 cm3 of 1 M acid and 40oC
C. 1g limestone powder, 100 cm3 acid of 1 M and 40oC
D. 1g limestone solid, 100 cm3 of acid of 1 M and 30oC

27. Which items correctly complete the following statement? A catalyst can act in a chemical
reaction to:
i. increase the equilibrium constant.
ii. lower the activation energy.
iii. decrease ∆E for the reaction.
iv. provide a new path for the reaction.
A. only I & II
B. only II & III
C. only III & IV
D. only II & IV

28. In the graph showing Maxwell Boltzman distribution of energy, ___________.
I. area under the curve must not change with increase in temperature.
II. area under the curve increases with increase in temperature.

III. area under the curve decreases with increase in temperature.
IV. with increase in temperature curve broadens and shifts to the right

hand side.

A. I,IV
B. II,IV
C. III,IV
D. IV

29. Arrhenius equation shows the variation of ___________________ with temperature?
A. Reaction rate
B. Rate constant
C. Energy of activation
D. Frequency factor

36

ANSWER PRACTICE MODULE CHAPTER 1 SK025

30. Consider the Arrhenius equation given below and mark the correct option.
= − /

A. Rate constant increases exponentially with increasing activation energy and
decreasing temperature.

B. Rate constant decreases exponentially with increasing activation energy and
decreasing temperature.

C. Rate constant increases exponentially with decreasing activation energy and
decreasing temperature.

D. Rate constant increases exponentially with decreasing activation energy
and increasing temperature

PSPM MODEL QUESTION

Worksheet 5

1 (a) The graph in FIGURE 1 represent the experiment of the reaction of A product.

(i) Based on FIGURE 1, determine the order of the reaction.
(ii) Determine the rate constant of the experiment.
Ans:

0 order reaction because graph [A] against time is a linear /straight line graph.
Rate constant, k= -gradient
k=0.021 M/min

37

ANSWER PRACTICE MODULE CHAPTER 1 SK025
(b) For the reaction P + Q → R, the following data were obtained from the experiment

Ans:

(c) Calculate the activation energy for a first order reaction, given that the rate constant
Ans: is 1.50 x 10-2 s -1 at 27 oC and 3.25 x 10-1 s -1 at 35 oC.

2 Define activation energy

The rate constant for the forward reaction of a reaction between hydrogen gas and

iodine gas is 138 L mol-1 s-1. The activation energy for the reaction is 165 kJ mol-1 at

700°C.

H2 (g) + I2(g) 2HI(g) ∆H = -98.48 kJmol-1

Sketch and label an energy profile diagram for this reaction.

Determine the activation energy and rate constant for the reverse reaction at 700°C.

(Assume Arrhenius factor, A is the same for both forward and reverse reactions).

38

ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans:

The minimum amount of energy required to initiate a chemical reaction

Ea (rev) = 165 + 9.48

= 174.48 kJmol-1

k = A e –Ea @ ln k = ln A – Ea
RT
RT

ln 138 = ln A – 165 x 103

8.314 x 973

A = 9.956 x 1010 J

ln k = ln A – Ea
RT

= 25.34 – 174.48 x 103
8.314 x 973

= 3.755
k = 42.75 L mol-1 s-1

3. (a) For a reaction : 4NH3(g) + 3O2(g) 2N2(g) + 6H2O(l)
nitrogen gas was formed at a rate of 0.72 mol L−1s−1.

i. Write the rate differential equation for the above reaction.

ii. Calculate the rate of water formation.

iii. State the relationship between rate of water formation and rate of

oxygen consumption.

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ANSWER PRACTICE MODULE CHAPTER 1 SK025

Ans: i.
ii.

iii. rate of water formation is double than the rate of oxygen consumption.
(b) A series of experiment was conducted to determine the rate of formation of

nitrogen gas, which is the product of the reaction between nitrogen monoxide
gas and hydrogen gas. The result obtained are tabulated as in TABLE 1:

Specify the order of reaction with respect to NO and H2. Hence write the rate
equation.

Ans: (b)

40

ANSWER PRACTICE MODULE CHAPTER 1 SK025

(c) The rate constant for a certain reaction is 2.52×10−5s−1at 190°C and
6.30×10−4s−1at 230°C. Calculate the activation energy for the reaction.

Ans:

Prepared by : Mdm Farhana bt Umanan
Checked by : Mdm Nur Zarifah Syazana bt Hamzah

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