1
MCQ (Single Correct Answer)

AIPMT 2003

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation, $$k = A \cdot {e^{ - E{}^ * /RT}}$$. Activation energy (E$$ * $$) of the reaction can be calculated by plotting
A
$$k\,\,vs\,\,T$$
B
$$k\,\,vs\,\,{1 \over {\log T}}$$
C
$$\log \,k\,\,vs\,\,{1 \over T}$$
D
$$\log \,k\,\,vs\,{1 \over {\log T}}$$

Explanation

Arrhenius equation k = $$A{e^{ - {{{E_a}} \over {RT}}}}$$

$$ \Rightarrow $$ log k = log A - $${{{{E_a}} \over {2.303RT}}}$$

Comparing it with equation of straight line i.e.,

y = mx + C

On plotting log k vs $${1 \over T}$$, we get a straight line, the slope indicates the value of activation energy.
2
MCQ (Single Correct Answer)

AIPMT 2002

2A $$ \to $$ B + C

It would be a zero order reaction when
A
the rate of reaction is proportional to square of concentration of A
B
The rate of reaction remains same at any concentration of A
C
the rate remains unchanged at any concentration of B and C
D
the rate of reaction doubles if concentrations of B is increased to double.

Explanation

2A $$ \to $$ B + C

If it is zero order reaction r = k [A]o.

i.e the rate remains same at any concentration of 'A'. i.e independent upon concentration of A.
3
MCQ (Single Correct Answer)

AIPMT 2001

For the reaction;

2N2O5 $$ \to $$ 4NO2 + O2 rate and rate constant are 1.02 $$ \times $$ 10$$-$$4 and 3.4 $$ \times $$ 10$$-$$5 sec$$-$$1 respectively, then concentration of N2O5 at that time will be
A
1.732
B
3
C
1.02 $$ \times $$ 10$$-$$4
D
3.4 $$ \times $$ 105

Explanation

For the reaction;

2N2O5 $$ \to $$ 4NO2 + O2

This is a first order reaction.

$$ \therefore $$ rate = k [N2O5] ;

[N2O5] = $${{rate} \over k}$$

= $${{1.02 \times {{10}^{ - 4}}} \over {3.4 \times {{10}^{ - 5}}}}$$

= 3
4
MCQ (Single Correct Answer)

AIPMT 2001

When a bio-chemical reaction is carried out in laboratory, outside the human body in absence of enzyme, then rate of reaction obtained is 10$$-$$6 times, the activation energy of reaction in the presence of enzyme is
A
6/RT
B
P is required
C
different from E$$a$$ obtained in laboratory
D
can't say anything.

Explanation

From Arrhenius equation, k = Ae –Ea/RT

The activation energy of the reaction in the presence of enzyme is different from Ea obtained in laboratory.

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