Question

$A \rightarrow B$ The rate constants of the above reaction at $200 K$ and $300 K$ are $0.03 \min ^{-1}$ and $0.05 \min ^{-1}$ respectively The activation energy for the reaction is __$J$ (Nearest integer) (Given $: \ln 10=2.3$ 

$R =8.3 \,JK ^{-1} \,mol ^{-1}$ 

$\log 5=0.70$ 

$\log 3=0.48$ 

$\log 2=0.30)$

Hint:

The activation energy can be calculated using the Arrhenius equation, where the ratio of rate constants at two different temperatures is related to the exponential factor involving the activation energy.

Answer 1

Correct Answer: 2520

The correct answer is 2520.
$\log \frac{K_{300}}{K_{200}}=\frac{E_a}{2.3\times 8.314}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
$\log \frac{0.05}{0.03}=\frac{Ea}{2.305\times 8.314}\times \left[\frac{1}{200}-\frac{1}{300}\right]$
$E_a=2519.88J\Rightarrow E_a=2520J$

Answer 2

We can use the Arrhenius equation to solve for the activation energy:

\[ \log \left( \frac{K_{300}}{K_{200}} \right) = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

Substituting the given values:

\[ \log \left( \frac{0.05}{0.03} \right) = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{200} - \frac{1}{300} \right) \]

Solving for \( E_a \):

\[ \log \left( \frac{0.05}{0.03} \right) = \log 5 - \log 3 = 0.70 - 0.48 = 0.22 \]

\[ 0.22 = \frac{E_a}{2.3 \times 8.314} \times \left( \frac{1}{200} - \frac{1}{300} \right) \]

\[ 0.22 = \frac{E_a}{2.3 \times 8.314} \times \left( \frac{1}{60000} \right) \]

Solving for \( E_a \):

\[ E_a = 2519.88 \, \text{J} \quad \Rightarrow \quad E_a \approx 2520 \, \text{J} \]

The activation energy (\( E_a \)) is calculated by rearranging the Arrhenius equation and solving for the unknown. This value is crucial in understanding the rate of a chemical reaction and the energy barrier that must be overcome for the reaction to proceed.