JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The rate constants for decomposition of acetaldehyde have been measured over the temperature range \(700-1000\, K\). The data has been analysed by plotting \(ln\, k\,\)vs\(\,\frac{10^{3}}{ T }\) graph. The value of activation energy for the reaction is \(......\,kJ mol ^{-1}\). (Nearest integer) (Given : \(R =8.31\, J \,K ^{-1} \,mol ^{-1}\) )
- A \(234\)
- B \(154\)
- C \(701\)
- D \(185\)
Answer & Solution
Correct Answer
(B) \(154\)
Step-by-step Solution
Detailed explanation
\(\ln k =\ln A -\frac{ Ea }{10^{3} RT } \times 10^{3}=\ln A+\frac{10^{3}}{ T }\left[-\frac{ Ea }{10^{3} RT }\right]\) From the graph \(\frac{- Ea }{10^{3} \times R }=-18.5\) \(Ea =153.735 \,k\,J / mol\) \(\sim 154\)
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