Discussion Forum

1)

 For a particular reaction , the rate constant becomes double on increasing temperature from  $27^{0} C$  to $37^{0}$C . Calculate the approximate activation energy

(in kcal  mol ^-1 R =2 cal mol^-1 K^-1)

Answer:

Option B

Explanation:

We know that

$\log\frac{k_{2}}{k_{1}}=\frac{E_{a}}{2.303 R}\times\frac{10}{93 \times1000}$

 $E_{a}=0.3010\times2\times93\times100\times2.303$

$E_{a}=5598.6cal\times2.303$

 $E_{a}=56\times2.303=12.89 kcal/mol$