1. Tardigrade
  2. Question
  3. Chemistry
  4. For reaction A arrow B, the rate constant k1=A1(e- Ea1 / RT) and for the reaction X arrow Y, the rate constant k2=A2(e- Ea2 / RT)⋅ If A1=109,A2=1010,Ea1=1200cal/mol and Ea2=1800cal/mol, then the temperature at which k1=k2 is (x/2 . 303)K. Find Y=(x/100) (Given; R=2cal/K-mol )

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Q. For reaction $A \rightarrow B,$ the rate constant $k_{1}=A_{1}\left(e^{- E_{a_{1}} / RT}\right)$ and for the reaction $X \rightarrow Y,$ the rate constant $k_{2}=A_{2}\left(e^{- E_{a_{2}} / RT}\right)\cdot $ If $A_{1}=10^{9},A_{2}=10^{10},E_{a_{1}}=1200cal/mol$ and $E_{a_{2}}=1800cal/mol,$ then the temperature at which $k_{1}=k_{2}$ is $\frac{x}{2 . 303}K.$ Find $Y=\frac{x}{100}$
(Given; $R=2cal/K-mol$ )

NTA AbhyasNTA Abhyas 2022

Solution:

$A \rightarrow B$
$K_{1}=A_{1}\times e^{\frac{- E_{a_{1}}}{RT}}$
$x \rightarrow y$
$K_{2}=A_{2}\times e^{\frac{- Ea_{2}}{RT}}$
if $k_{1}=k_{2}$
$10^{9}e^{\frac{- 1200}{RT}}=10^{10}e^{\frac{- 1800}{RT}}$
$\frac{- 1200}{RT}=\frac{- 1800}{RT}+ln10$
$T=\frac{300}{2 . 303}K$