Question

How much faster would a reaction proceed at ${25}^{\circ }C$ than at $0^{\circ }C$, if the activation energy is $65\text{kJ}$? (Given: $R=8.3{\text{J K}}^{-1}{\text{mol}}^{-1},{10}^{1.043}=11$)

• A. $2$ times
• B. $5$ times
• C. $11$ times
• D. $16$ times

Answer 1

The correct option is C $11$ times

Given,
$T_2={25}^{\circ }C=25+273=298\text{K}{T}_1=0^{\circ }C=0+273=273\text{K}$
By using <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Arrhenius equation:
$\log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\frac{(T_2-T_1)}{T_1{T}_2}$
Substituting the values, we get
$\log \frac{k_2}{k_1}=\frac{65\times {10}^3\times (298-273)}{2.303\times 8.3\times 298\times 273}\Rightarrow \log \frac{k_2}{k_1}=1.043\Rightarrow k_2=11\times k_1$