# Topic: Unit VII: Chemical Kinetics (Test 1)

Topic: Unit VII: Chemical Kinetics
Q.1
For a first order reaction A → B the rate constant is x min1 . 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
Explaination / Solution:

In this case

k = x min−1 and [A0] = 0.01M = 1 × 10−2M

t = 1 hour = 60 min

[A]= 1 × 10−2 ( e−60x)

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Q.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
Explaination / Solution:

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Q.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
A.
B.
C.
D. both (b) and (c)
Explaination / Solution:

k = A e –(Ea/RT)

ln k = ln A – (Ea/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

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Q.4
For a first order reaction A → product with initial concentration x mol L1 , has a half life period of 2.5 hours . For the same reaction with initial concentration (x/2) mol L1 the half life is
A. (2.5 × 2) hours
B. (2.5 / 2) hours
C. 2.5 hours
D. Without knowing the rate constant, t1/2 cannot be determined from the given data
Explaination / Solution:

For a first order reaction

t1/2 = 0.693/k

t1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)

t1/2 = 2.5 hrs

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Q.5

For the reaction, 2NH3 → N2 + 3H2

then the relation between k1, k2 and k3 is
A. k1 = k2 = k3
B. k1 = 3k2 = 2k3
C. 1.5 k1 = 3 k2 = k3
D. 2 k1 = k2 = 3 k3
Explaination / Solution:

(3/2)k1 =  3k2 = k3

1.5 k1 = 3k2 = k3

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Q.6
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
Explaination / 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.

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Q.7
For a reaction Rate = k[acetone]3/2 acetone then unit of rate constant and rate of reaction respectively is
A. (mol L-1s-1), (mol-1/2 L1/2 s-1)
B. (mol-1/2 L1/2s-1), (mol L-1 s-1)
C. (mol1/2 L1/2s-1), (mol L-1 s-1)
D. (mol Ls-1), (mol1/2 L1/2 s-1)
Explaination / Solution:

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Q.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
Explaination / Solution:

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

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Q.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 Ea = 0

(iii) rate constant k is equal to collision frequency A if Ea = °

(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)
Explaination / Solution:

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

k = A e – (Ea/RT)

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

k = A e0

k = A

ln k = ln A – (Ea /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.

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Q.10
In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectivelyx kJ mol1 and y kJ mol1 . Therefore , the energy of activation in the backward direction is
A. ( y x ) kJ mol1
B. ( x + y )J mol1
C. ( x y ) kJ mol1
D. ( x + y ) × 103 J mol1