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(C) The rate of reaction $k$ is inversely proportional to the time taken $t$, so $k \propto 1/t$.Using the Arrhenius equation: $\ln \left(\frac{k_2}{k

Vedclass · Accepted Answer

$0.693 R \frac{T_1 T_2}{T_2-T_1}$

Vedclass · Answer

(C) The rate of reaction $k$ is inversely proportional to the time taken $t$, so $k \propto 1/t$.Using the Arrhenius equation: $\ln \left(\frac{k_2}{k_1}ight) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.Given $t_1 = 4.0 	ext{ min}$ at $T_1$ and $t_2 = 8.0 	ext{ min}$ at $T_2$, we have $k_1 = 1/4$ and $k_2 = 1/8$.Substituting these values: $\ln \left(\frac{1/8}{1/4}ight) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.$\ln(0.5) = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.Since $\ln(0.5) = -\ln(2) \approx -0.693$, we get $-0.693 = \frac{E_a}{R} \left[\frac{T_1 - T_2}{T_1 T_2}ight]$.Rearranging for $E_a$: $E_a = -0.693 R \left[\frac{T_1 T_2}{T_1 - T_2}ight] = 0.693 R \left[\frac{T_1 T_2}{T_2 - T_1}ight]$.Note: The provided options suggest a specific sign convention; based on the derivation, the correct expression is $0.693 R \frac{T_1 T_2}{T_2 - T_1}$.

Similar Questions

The rate of reaction increases with a rise in temperature because of:

$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
The curves $M$ and $N$ represent the variation of energy with reaction coordinate for the reaction in absence and presence of catalyst.
Which value represents the activation energy $(E_a)$ for the backward reaction in the presence of catalyst?

In the presence of a catalyst,the heat evolved or absorbed during the reaction . . . . . . .

Which of the following graphs for $\ln k$ versus $\frac{1}{T}$ is correct?

If for a hypothetical reaction,$E_a = 0$ at $273 \ K$,then find the ratio of the rate constants at $383 \ K$ and $273 \ K$.

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