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Questions:

Multiple Choice Question

24 x 1 = 24

This reaction follows first order kinetics. The rate constant at particular temperature is 2.303 x 10^-2 hour^-1. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (log 2 = 0.3010)

(a)

0.125 M

(b)

0.215 M

(c)

0.25 x 2.303 M

(d)

0.05 M

After 2 hours, a radioactive substance becomes \(\left( \frac { 1 }{ 16 } \right) ^{ th }\) of original amount Then the half life (in min) is _______.

(a)

60 minutes

(b)

120 minutes

(c)

30 minutes

(d)

15 minutes

The correct difference between first and second order reactions is that________.

(a)

A first order reaction can be catalysed; a second order reaction cannot be catalysed.

(b)

The half life of a first order reaction does not depend on [A₀]; the half life of a second order reaction does depend on [A₀].

(c)

The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.

(d)

The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations.

The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to

(a)

\(\left( \frac { 1 }{ 2 } \right) g\)

(b)

\(\left( \frac { 1 }{ 4 } \right) g\)

(c)

\(\left( \frac { 1 }{ 8 } \right) g\)

(d)

\(\left( \frac { 1 }{ 16 } \right) g\)

If 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in_______.

(a)

20 minutes

(b)

30 minutes

(c)

35 minutes

(d)

75 minutes

In a homogeneous reaction A⟶B+C+D, the initial pressure was P₀ and after time t it was P expression for rate constant in terms of P₀, P and t will be _____.

(a)

\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-P } \right) \)

(b)

\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { { 2P }_{ 0 } }{ { P }_{ 0 }-P } \right) \)

(c)

\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { 3{ P }_{ 0 }-P }{ 2P_{ 0 } } \right) \)

(d)

\(k=\left( \frac { 2.303 }{ t } \right) \log\left( \frac { 2{ P }_{ 0 } }{ { 3P }_{ 0 }-2P } \right) \)

If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is______.

(a)

Zero

(b)

one

(c)

Fraction

(d)

none

During the decomposition of H₂O₂ to give dioxygen, 48 g O₂ is formed per minute at certain point of time. The rate of formation of water at this point is

(a)

0.75 mol min⁻¹

(b)

1.5 mol min⁻¹

(c)

2.25 mol min⁻¹

(d)

3.0 mol min⁻¹

For the reaction \({ N }_{ 2 }{ O }_{ 5 }\left( g \right) \longrightarrow { 2NO }_{ 2 }\left( g \right) +\frac { 1 }{ 2 } { O }_{ 2 }\left( g \right) \) value of rate of disappearance of N₂O₅is given as 6.5 x 10^-2 mol L^-1s^-1. The rate of formation of NO₂ and O₂ is given respectively as_____.

(a)

(3.25 x 10^-2 mol L^-1s^-1) and (1.3 x 10^-2 mol L^-1s^-1)

(b)

(1.3 x 10^-2 mol L^-1s^-1) and (3.25 x 10^-2 mol L^-1s^-1)

(c)

(1.3 x 10^-1mol L^-1s^-1) and (3.25 x 10^-2 mol L^-1s^-1)

(d)

None of these

10.

The rate constant of a reaction is 5.8 x 10^-2 S^-1 The order of the reaction is ______.

(a)

First order

(b)

zero order

(c)

Second order

(d)

Third order

11.

Predict the rate law of the following reaction based on the data given below
2A+B⟶C+3D

Reaction number	[A] (min)	[B] (min)	Initial rate (M s^-1)
1	0.1	0.1	x
2	0.2	0.1	2x
3	0.1	0.2	4x
4	0.2	0.2	8x

(a)

rate = k[A]²[B]

(b)

rate = k[A] [B]²

(c)

rate = k[A] [B]

(d)

rate = k[A]^1/2[B]^3/2

12.

In a first order reaction x ⟶ y; if k is the rate constant and the initial concentration of the reactant x is 0.1M, then, the half life is_____.

(a)

\(\left( \frac { \log2 }{ k } \right) \)

(b)

\(\left( \frac { 0.693 }{ (0.1)k } \right) \)

(c)

\(\left( \frac { In2 }{ k } \right) \)

(d)

none of these

13.

For a first order reaction, the rate constant is 6.909 min^-1the time taken for 75% conversion in minutes is _______.

(a)

\(\left( \frac { 3 }{ 2 } \right) { \log 2 }\)

(b)

\(\left( \frac { 2 }{ 3 } \right) \log2\)

(c)

\(\left( \frac { 3 }{ 2 } \right) \log\left( \frac { 3 }{ 4 } \right) \)

(d)

\(\left( \frac { 2 }{ 3 } \right) \log\left( \frac { 4 }{ 3 } \right) \)

14.

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R = 8.314 JK^-1 mol^-1)

(a)

234.65 kJ mol^-1

(b)

434.65 kJ mol^-1

(c)

2.305 kJ mol^-1

(d)

334.65 J mol^-1

15.

In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively −x kJ mol^-1 and y kJ mol^-1. Therefore, the energy of activation in the backward direction is _______.

(a)

(y-x) kJ mol^-1

(b)

(x+y) J mol^-1

(c)

(x-y) KJ mol^-1

(d)

(x+y) x 10³J mol^-1

16.

Consider the following statements:
(i) increase in concentration of the reactant increases the rate of a zero order reaction.
(ii) rate constant k is equal to collision frequency A if E_a = 0
(iii) rate constant k is equal to collision frequency A if E_a = ∞
(iv) a plot of ln (k) vs T is a straight line.
(v) a plot of ln (k) vs \(\left( \frac { 1 }{ T } \right) \) is a straight line with a positive slope.
Correct statements are

(a)

(ii) only

(b)

(ii) and (iv)

(c)

(ii) and (v)

(d)

(i), (ii) and (v)

17.

The addition of a catalyst during a chemical reaction alters which of the following quantities?

(a)

Enthalpy

(b)

Activation energy

(c)

Entropy

(d)

Internal energy

18.

For a reaction Rate = k[acetone]^3/2 then unit of rate constant and rate of reaction respectively is _______.

(a)

(mol L^-1 S^-1),(mol^1/2L^1/2S^-1)

(b)

(mol^-1/2L^1/2s^-1),(mol L^-1s^-1)

(c)

(mol^1/2L^1/2s^-1),(mol L^-1s^-1)

(d)

(mol L s^-1),(mol^1/2L^1/2s)

19.

The decomposition of phosphine (PH₃) on tungsten at low pressure is a first order reaction. It is because the _____.

(a)

rate is proportional to the surface coverage

(b)

rate is inversely proportional to the surface coverage

(c)

rate is independent of the surface coverage

(d)

rate of decomposition is slow

20.

For the reaction, 2NH₃ ⟶ N₂ + 3H₂, if \(\frac { -d[NH_{ 3 }] }{ dt } \) = k₁[NH₃], \(\frac { d[N_{ 2 }] }{ dt } =k_{ 2 }[NH_{ 3 }],\frac { d[{ H }_{ 2 }] }{ dt } \)= k₃[NH₃] then the relation between k₁, k₂ and k₃ is _________.

(a)

k₁= k₂= k₃

(b)

k₁= 3k₂= 2k₃

(c)

1.5k₁= 3k₂= k₃

(d)

2k₁= k₂= 3k₃

21.

For a first order reaction A ⟶ product with initial concentration x mol L^-1, has a half life period of 2.5 hours. For the same reaction with initial concentration \(\left( \frac { x }{ 2 } \right) \) mol L^-1 the half life is

(a)

(2.5 x 2) hours

(b)

\(\left( \frac { 2.5 }{ 2 } \right) \) hours

(c)

2.5 hours

(d)

Without knowing the rate constant, t_1/2 cannot be determined from the given data

22.

Among the following graphs showing variation of rate constant with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is _______.

(a)

(b)

(c)

(d)

both (b) and (c)

23.

A zero order reaction X ⟶ Product, with an initial concentration 0.02M has a half life of 10 min. if one starts with concentration 0.04M, then the half life is

(a)

10 s

(b)

5 min

(c)

20 min

(d)

cannot be predicted using the given information

24.

For a first order reaction A ⟶ B the rate constant is x min⁻¹. If the initial concentration of A is 0.01M, the concentration of A after one hour is given by the expression.

(a)

001. e^−x

(b)

1 x 10^-2(1-e^-60x)

(c)

(1 x 10^-2)e^-60x

(d)

none of these

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Answers Key:

Multiple Choice Question

24 x 1 = 24

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
2.303 x 10^-2 hour^-1 = \(\frac { 2.303 }{ 1806 min } log\frac { \left[ { 0.25 }_{ } \right] }{ \left[ A \right] } \)
\(=\left(\frac{2.303 \times 10^{-2} hour^{-1 }\times 1806 min}{2.303}\right) = \log \left(\frac{0.25}{A}\right) \)
\(=\left(\frac{ 1806 \times 10 ^{-2}}{60}\right) = \log \left(\frac{0.25}{A}\right) \)
\(= 0.301 = \log \left(\frac{0.25}{A}\right) \)
\(=\log2 = \log \left(\frac{0.25}{A}\right) \)
\(2 = \log \left(\frac{0.25}{A}\right) \)
\([A] = \log \left(\frac{0.25}{2}\right) = 0.125 M\)

\({ t }_{ 1/2 }=\frac { 0.6932 }{ k } \)
For a second order reaction
\(\mathrm{t}_{1 / 2}=\frac{2^{\mathrm{n}-1}-1}{(\mathrm{n}-1) \mathrm{k} \cdot\left[\mathrm{A}_{0}\right]^{\mathrm{n}-1}}\)
n = 2
\(\mathrm{t}_{1 / 2}=\frac{2^{\mathrm{2}-1}-1}{(\mathrm{n}-1) \mathrm{k} \cdot\left[\mathrm{A}_{0}\right]^{\mathrm{2}-1}}\)
\(\mathrm{t}_{1 / 2} =\frac{1}{ \mathrm{k}\left[\mathrm{A}_{0}\right]}\)

In 140days ⇒ initial concentration reduced to (1/2) g
In 280 days ⇒ initial concentration reduced to (1/4) g
In 420 days ⇒ initial concentration reduced to (1/8) g
In 560 days ⇒ initial concentration reduced to (1/16) g

t_75% = 2t_50%
t_50% = (t_75%/2) = (60/2) = 30 minutes

	A	⟶	B	C	D
Initial rate (M s^-1)	a		0	0	0
Reaction number	x	-	-	-	-
After time t	(a - x)		x	x	x
Total number of moles	= (a + 2x)

t_1/2 α \(\frac{1}{[A_{0}]^{n-1}}\)...(1)
If [A₀ = 2[A₀]; then t_1/2 = 2t_1/2
2t_1/2 α \(\frac{1}{[2A_{0}]^{n-1}}\)...(2)
(2)/(1) = \(2= \frac{1}{[2A_{0}]^{n-1}} \times \frac{1}{[A_{0}]^{n-1}}\)
\(2= \frac {[2A_{0}]^{n-1}} {[A_{0}]^{n-1}}\)
\(2= \frac {1} {2}^{n-1}\)
2 = (2^-1)^n-1
2¹ = (2^-n+1)
n = 0

(d)
3.0 mol min⁻¹

Rate \(=\frac { d\left[ { N_2O_5 } \right] }{ dt } =\frac { 1 }{ 2 } =\frac { d\left[ { NO}_{ 2 } \right] }{ dt } =\frac { 2d\left[ { O }_{ 2 } \right] }{ dt } \)
Given that \(=\frac { d\left[ { N_2O_5 } \right] }{ dt } \) =6.5 x 10^-2 mol L^-1 s^-1
\(=\frac { d\left[ { NO}_{ 2 } \right] }{ dt }\) =2 x 6.5 x 10^-2= 1.3 x 10^-1mol L^-1 s^-1
\(=\frac { d\left[ { O}_{ 2 } \right] }{ dt }\)\(=\frac{6.5 \times 10^{-2}}{2}\) = 3.25 x 10^-2 mol L^-1s^-1

10.

(a)
First order

11.

rate₁ = k[0.1]ⁿ [0.1]^m ....(1)
rate₂ = k[0.2]ⁿ [0.1]^m ....(2)
(2) (1)
\(=\frac{2x}{x} = \frac{k[0.2]^n [0.1]^m}{k[0.1]^n [0.1]^m}\)
\(= \frac{2x}{x} = 2^n\)
n = 1
rate₃ = k[0.1]ⁿ [0.2]^m ....(3)
rate₄ = k[0.2]ⁿ [0.2]^m ....(4)
\(=\frac{8x}{2x} = \frac{k[0.2]^n [0.2]^m}{k[0.2]^n [0.1]^m}\)
= 8/2 = 2m
m = 2
rate = k[A] [B]²

12.

\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 0.1: [A] = 0.05
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln\left[ \frac { { 0.1 } }{ 0.05 }\right] \)
\(k= \left[ \frac { 1 }{ t _{1/2}} \right] ln (2)\)
t_1/2 = In(2)/K

13.

\(k=\frac { 2.303 }{ t } \log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
[A₀] = 100: [A] = 25
\(6.909=\frac { 2.303 }{ t } \log\frac { \left[ {100 } \right] }{ \left[ 25\right] } \)
\(t =\frac { 2.303 }{ 6.909 } \log(4)\)
\(t =\frac { 1 }{ 3 } \log(2^2)\)
\(= \left( \frac { 2 }{ 3 } \right) \log2\)

14.

T₁= 200 K ; k = k₁
T₂ =400 K ; k₁= k₂= 2k₁
log\(\frac {{ k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right] \)
log \( [\frac {{2 k }_{ 1 } }{ { k }_{ 1 } }]\)
\(= \left[ \frac { E_{a} }{ 2.303\times 8.314 JK^{-1} mol^{-1} } \right] \)
\(= \left[ \frac { 400 k - 200K }{200 k \times 400 k } \right] \)
\(E_{a} = \frac{ 0.3010 \times 2.303 \times 8.314 JK^{-1}mol^{-1} \times 200K \times 400 K}{200 K}\)
E_a = 2305 J mol^-1
E_a = 2.305 kJ mol^-1

15.

16.

(rate constant K is equal to collision frequency A if E_a = 0)
In zero order reactions, increase in the concentration of reactant does not alter the rate.
So statement (i) is wrong.
\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
if E_a = 0 so, statement (ii) is correct, and statement (iii) is wrong
k = Ae⁰
k = A
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope
So statement (iv) and (v) are wrong.

17.

A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

18.

Rate = k[A]ⁿ
Rate = \(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\)
unit of rate = \(\frac{mol L^{-1}}{s}\)=mol L^-1/s^-1
unit of rate constant
\(=\frac{ (mol{ L }^{ -1 }{ S }^{ -1 }) }{ ({ mol }{ L }^{ -1 })^n } \)
= mol^1-nL^n-1s^-1
in the case
rate = k [Acetone]^3/2
n = 3/2
= mol^1-(3/2)L^(3/2)-1s^-1
(mol^-(1/2)L^(1/2)s^-1).

19.

Given:
At low pressure the reaction follows first order therefore,
Rate α [reactant]¹
Rate α (surface area)
At high pressure due to the complete coverage of surface area, the reaction follows zero order.
Rate α [reactant]⁰
Therefore the rate is independent of surface area.

20.

\(Rate=\frac { -1 }{ 2 } \frac { d[NH_3] }{ dt } \)
\(\frac { -d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { H}_{ 2 } \right] }{ dt } \)
\(=\frac { 1 }{ 2 } { k_1\left[ { NH}_{ 3 } \right] } = { k_2 \left[ { NH}_{ 3 } \right] }=\frac { 1 }{ 3 } { k_3\left[ { NH}_{ 3 } \right] } \)
[3/2] k₁ = 3k₂ = k₃
1.5k₁= 3k₂= k₃

21.

For a first order reaction
t_1/2 = \(\frac { 0.693 }{ { k }}\)
t_1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
t_1/2 = 2.5 hrs

22.

\(k=A{ e }^{ -\left( \frac { { E }_{ a } }{ RT } \right) }\)
In k = In A - \({ \left( \frac { { E }_{ a } }{ R } \right) }\) \(\left( \frac { 1 }{ T } \right) \)
This equation is of the form of a straight line y = mx+c
A plot of In k Vs \(\left( \frac { 1 }{ T } \right) \) gives a straight line with a negative slope.

23.

for n ≠ 1 t_1/2 = \(\frac{2^{n-1} -1}{(n- 1) k[A_{0}]^{-1}}\)
for n = 0; t_1/2 = \(\frac{1}{2 k[A_{0}]^{-1}}\)
t_1/2 = \(\frac{[A_{0}]}{2 k}\)
t_1/2 α [A₀] ...(1)
Given [A₀] = 0.002 M; t_1/2 = 10 min
[A₀] = 0.04M; t_1/2 = ?
Substitute in (1)
10 min α 0.02 M...(2)
t_1/2 α 0.04M ....(3)
(3)(2)
⇒ t_1/2 / 10 min
= 0.04 M/0.02 M
t_1/2 = 2 x 10 min = 20 min

24.

\(k=\frac { 2.303 }{ t } log\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(k=\frac { 1 }{ t } ln \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \)
\(e^{kt}=\frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] }\)
[A] = [A₀] e^kt
In this case
k = x min^-1 and [A₀] = 0.01 M = 1 x 10^-2M
t = 1 hour = 60 min
[A] = (1 x 10^-2)e^-60x