The temperature at which the rate constants o | vedclass

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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}}

Vedclass · Accepted Answer

$1303$

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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 	imes 2.303 = \frac{6000}{T}$$T = \frac{6000}{4.606} \approx 1302.64 \ K$Rounding to the nearest integer, we get $T = 1303 \ K$.

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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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: $\ln 10 = 2.303$