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VITEEE 2016 Paper

Section: Mathematics

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Question : 74 of 125

Marks: +1, -0

The following data were obtained for a given reaction at 300 K.

Reaction	Energy of activation (kJ $\mathrm{mol}^{-1}$ )
(i) uncatalysed	76
(ii) catalysed	57

The factor by which rate of catalysed reaction is increased, is

21
2100
2000
1200

Solution:

Using Arrhenius equation,

K =

A \cdot e^{-\frac{E_a}{Rt}}

, we get

log k = log A -

\frac{E_a}{2.303\,RT}

∴ log

k_1

= log A -

\frac{E_{a(1)}}{2.303\,RT}

... (i)

and log

k_2

= log A -

\frac{E_{a(2)}}{2.303\,RT}

... (ii)

or log

\frac{k_2}{k_1}

=

\frac{1}{2.303\,RT}

\left[ E_{a(1)} - E_{a(2)} \right]

(from (i) and (ii))

=

\frac{1}{2.303 \times 8.314 \times 300}

(76000 - 57000)

or log

\frac{k_2}{k_1}

=

\frac{19000}{2.303 \times 8.314 \times 300}

=

\frac{190}{6.9 \times 8.314}

or

\frac{k_2}{k_1}

= 2000 [taking antilog]

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