For the reaction following data is givenA→B;k1=1015exp⁡−2000TC→D;k2=1014exp⁡−1000TAt what temperature, k1 and k2 will be same ? _____Round off to nearest integer.

k1 = k21015e−2000T=1014e−1000TOr 15−2000T×2.303=14−1000T×2.303Or 1000T×2.303=1∴ T = 434.22K

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For the reaction following data is given
$\begin{array}{l} A \to B; k_{1} = 10^{15} \exp (\frac{- 2000}{T}) \\ C \to D; k_{2} = 10^{14} \exp (\frac{- 1000}{T}) \end{array}$
At what temperature, k₁ and k₂ will be same ? _____Round off to nearest integer.

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answer is 434.

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Detailed Solution

k₁ = k₂
$10^{15} e^{(\frac{- 2000}{T})} = 10^{14} e^{(\frac{- 1000}{T})}$
Or $15 - \frac{2000}{T \times 2 .303} = 14 - \frac{1000}{T \times 2 .303}$
Or $\frac{1000}{T \times 2 .303} = 1$
$∴$ T = 434.22K

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