(C) Equating the two rate constants: $k_1 = k_2$$10^6 e^{\frac{-30000}{T}} = 10^4 e^{\frac{-24000}{T}}$Divide both sides by $10^4 e^{\frac{-30000}{T}}

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(C) Equating the two rate constants: $k_1 = k_2$$10^6 e^{\frac{-30000}{T}} = 10^4 e^{\frac{-24000}{T}}$Divide both sides by $10^4 e^{\frac{-30000}{T}}$:$10^2 = e^{\frac{-24000}{T} - (\frac{-30000}{T})}$$100 = e^{\frac{6000}{T}}$Taking natural logarithm on both sides:$\ln(100) = \frac{6000}{T}$$2 \ln(10) = \frac{6000}{T}$$2 \times 2.303 = \frac{6000}{T}$$T = \frac{6000}{4.606} \approx 1302.64 \ K$Rounding to the nearest integer, we get $T = 1303 \ K$.

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

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The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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What do $Z_{AB}$ and $P$ indicate in the context of collision theory?

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The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$Given: $\ln 10 = 2.303$

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Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

The rate of the process doubles with every $10 \ K$ increase in temperature. When the temperature is increased from $303 \ K$ to $353 \ K$,how much will the rate of the process increase?

For the reaction $A + B \rightarrow C + D$,$\Delta H = -20 \ kJ \ mol^{-1}$. The activation energy for the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the reverse reaction is ......... $kJ \ mol^{-1}$.

What do $Z_{AB}$ and $P$ indicate in the context of collision theory?

In the Arrhenius equation,the pre-exponential factor represents ...

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The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$