When a biochemical reaction is carried out in laboratory in the absence of enzyme then rate of reaction…
When a biochemical reaction is carried out in laboratory in the absence of enzyme then rate of reaction obtained is $10^{-6}$ times, then activation energy of reaction in the presence of enzyme is
$\frac{6}{\mathrm{RT}}$
different from $\mathrm{E}_{\mathrm{a}}$ obtained in laboratory
$\mathrm{P}$ is required
can't say anything
Solution
The presence of enzyme (catalyst) increases the speed of reaction by lowering the energy barrier, i.e., a new path is followed with lower activation energy.
Here $\mathrm{E}_{\mathrm{T}}$ is the threshold energy. $E_{a}$ and $E_{a l}$ is energy of activation of reaction in absence and presence of catalyst respectively.
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