Chemistry : Chemical Kinetics: Multiple choice questions with answers, Solution and Explanation

**EVALUATION**

1. 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) 0.01 e^{−x}

b) 1 x10^{-2} (1-e^{-60x})

**(c) (1x10 ^{-2}) e^{-60x}**

d) none of these

**Solution**

In this case

k = x min^{−1} and [A_{0}]
= 0.01M = 1 × 10^{−2}M

t = 1 hour = 60 min

[A]= 1 × 10^{−2} ( e^{−60x})

2. 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

**Solution**

3. 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

d) both (b) and (c)

**Ans: (b)**

**Solution**

k = A e ^{–(Ea/RT)}

ln k = ln A – (E_{a}/R)
(1/T)

this equation is in the
form of a straight line equatoion

y = c + m x

a plot of lnk vs 1/T is
a straight line with negative slope

4. 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 (x/2) mol L^{−}^{1} the half life is

a) (2.5 × 2) hours

b) (2.5 / 2) hours

c) 2.5 hours

**d) Without knowing the rate constant, t _{1/2} cannot be
determined from the given data **

**Solution**

For a first order reaction

t_{1/2} = 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

5. For the reaction, 2NH_{3} → N_{2} + 3H_{2}

then the relation between k_{1}, k_{2} and k_{3}
is

a) k_{1} = k_{2} = k_{3}

b) k_{1} = 3k_{2} = 2k_{3}

**c) 1.5 k _{1} = 3
k_{2} = k_{3} **

d) 2 k_{1} = k_{2} = 3 k_{3}

**Solution**

(3/2)k_{1} = 3k_{2}
= k_{3}

1.5 k_{1} = 3k_{2} = k_{3}

6. The decomposition of phosphine (PH_{3}) 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

**Solution**

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.

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

a) (mol L^{-1}s^{-1}),
(mol^{-1/2} L^{1/2} s^{-1})

**b) (mol ^{-1/2} L^{1/2}s^{-1}), (mol L^{-1}
s^{-1})**

c) (mol^{1/2} L^{1/2}s^{-1}),
(mol L^{-1} s^{-1})

d) (mol Ls^{-1}),
(mol^{1/2} L^{1/2} s^{-1})

**Solution**

8. 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

**Solution**

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

9. 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 1/T
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)

**Solution**

In zero order reactions, increase in the concentration of reactant
does not alter the rate. So statement (i) is wrong.

k = A e ^{– (E}^{a}^{/RT)}

if Ea = 0 so, statement
(ii) is correct, and statement (iii) is wrong

k = A e^{0}

k = A

ln k = ln A – (E_{a} /R) (1/ T)

this equation is in the form of a straight line equatoion

y = c + m x

a plot of lnk vs 1/T is a straight line with negative slope

so statements (iv) and (v) are wrong.

10. 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 ***)*** ***×*** ***103 J mol**

**Solution**

(x+y) kJmol^{-1}

(x+y) 10^{3} Jmol^{-1}

11. 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}K^{−}^{1}

b) 434.65 kJ mol^{−}^{1}K^{−}^{1}

**c) 434.65 J mol**^{−}^{1}**K**^{−}^{1}

d) 334.65 J mol^{−}^{1}K^{−}^{1}

**Solution**

T_{1} = 200K ; k = k_{1}

T_{2} = 400K ; k = k_{2}
= 2k_{1}

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

a) 0.125M

**b) 0.215M**

c) 0.25 × 2.303M

d) 0.05M

**Solution**

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

**Ans: (b)**

**Solution**

14. 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

**Ans: (c)**

**Solution**

15. Predict the rate law
of the following reaction
based on the data given below

2A + B → C + 3D

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}

**Solution**

16. Assertion: rate of reaction doubles when the concentration of
the reactant is doubles if it is a first order reaction.

Reason: rate constant
also doubles

a) Both assertion and reason are true and reason is the correct
explanation of assertion.

b) Both assertion and reason are true but reason is not the
correct explanation of assertion.

**c) Assertion is true but
reason is false.**

d) Both assertion and reason
are false.

**Solution**

For a first reaction, If
the concentration of reactant is doubled, then the rate of reaction also
doubled.

Rate constant is
independent of concentration and is a constant at a constant temperature,

17. The rate constant of
a reaction is 5.8 × 10^{−2} s^{−1} . The order of the
reaction is

**a) First order **

b) zero order

c) Second order

d) Third order

**Solution**

The unit of rate
constant is s-1 and it indicates that the reaction is first order.

18. For the reaction N_{2}O_{5}(g) → 2NO_{2}
(g) + ½ O_{2}(g), the value of rate of disappearance of N_{2}O_{5}
is given as 6.5 × 10^{-2} mol L^{-1}
s^{-1}. The rate of formation of NO_{2} and O_{2} is
given _{ }respectively as_{}

a) ( 3.25 ×
10^{−}^{2} mol L^{−}^{1}s^{−}^{1} ) and (1.3 × 10^{−}^{2} mol L^{−}^{1}s^{−}^{1 })

b) (1.3 ×
10^{−}^{2} mol L^{−}^{1}s^{−}^{1} ) and ( 3.25 × 10^{−}^{2} mol L^{−}^{1}s^{−}^{1} )

**c) (****1.3 ****×**** 10**^{−}^{1}** mol L**^{−}^{1}**s**^{−}^{1}** ****)**** and ****(**** 3.25 ****×**** 10**^{−}^{2}** mol L**^{−}^{1}**s**^{−}^{1}** ****)**

d) None of these

**Solution**

19. During the decomposition of H_{2}O_{2} to give
dioxygen, 48 g O_{2} 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}

**Solution**

no of moles of oxygen = (48/32) =
1.5 mol

∴
rate of formation of oxygen = 2 ×1.5
= 3 mol min^{-1}

20. 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

**Solution**

For a first order
reaction t_{1/2} is independent of initial concentration .i.e., ∴
n ≠ 1; for such cases

21. In a homogeneous reaction A →B + C
+ D , the initial pressure was P_{0} and
after time t it was P. expression for rate
constant in terms of P_{0}, P and t will be

**Ans: (a)**

**Solution**

22. 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

**Solution**

t_{75%} = 2t_{50%}

t_{50%} = (t_{75% }/
2) = (60/2) = 30 min

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

a) (1/2)g

b) (1/4)g

c) (1/8)g

**d) (1/16)g**

**Solution**

In 140 days ⟹ 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

24. 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 [A _{0}]; the half life of a
second order reaction does depend on [A_{0}].**

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.

**Solution**

For a first order
reaction

t_{1/2} =
0.6932/k

For a second order
reaction

25. After 2 hours, a
radioactive substance becomes (1/16)^{th} of original amount. Then the
half life ( in min) is

a) 60 minutes

b) 120 minutes

**c) 30 minutes **

d) 15 minutes

**Solution**

4t_{1/2} = 2
hours

t_{1/2} = 30 min

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