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

**Answer : ****Option C**

**Explaination / 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})

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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

**Answer : ****Option C**

**Explaination / Solution: **

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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)

**Answer : ****Option B**

**Explaination / 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

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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, t1/2 cannot be determined from the given data

**Answer : ****Option D**

**Explaination / 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

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

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

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

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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

**Answer : ****Option C**

**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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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})

**Answer : ****Option B**

**Explaination / Solution: **

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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

**Answer : ****Option B**

**Explaination / Solution: **

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

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

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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.

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 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.

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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^{−1}

**Answer : ****Option D**

**Explaination / Solution: **

(x+y) kJmol^{-1}

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

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