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The rate of a certain biochemical reaction at physiological temperature (T) 300 K occurs 10^{\9} times faster with enzyme than without. The Change in the activation energy upon adding enzyme is take R=8.3

Option: 1

-23410.6


Option: 2

-3178.9


Option: 3

-51610.23


Option: 4

+32316.56


Answers (1)

best_answer

\mathrm{\begin{aligned} & K=A e^{-E / R T} \ldots(i) \\ & 10^g K=A e^{-E c R T} \ldots(i i) \end{aligned}}

On dividing eq (ii) by (i)

\mathrm{10^9=e^{\left(E-E_C\right) / R T}}

On taking logarithm both sides 

\mathrm{\begin{aligned} & 9 \ln 10=\frac{(E-E C)}{R T} \\ & 2.303 \times 9=\frac{(E-E C)}{R T} \\ & E-E_c=2.303 \times(9 R T) \\ & \begin{aligned} \Delta E_a=E_c-E & =-2.303 \times 9 \times 8.3 \times 300 \\ & =-51610.23 \end{aligned} \end{aligned}}

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