TS EAMCET · Chemistry · Chemical Kinetics
The graph obtained between \(\ln \mathrm{k}(\mathrm{k}=\) Rate constant \()\) on \(y\)-axis \(1 / T\) on \(x\)-axis is a straight line. The slope of it is \(-4 \times 10^4 \mathrm{k}\). The activation energy of the reaction (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) is \[ \left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \]
- A \(166\)
- B \(332\)
- C \(765\)
- D \(382\)
Answer & Solution
Correct Answer
(B) \(332\)
Step-by-step Solution
Detailed explanation
The equation for the corresponding graph is :- \[ \ln \mathrm{K}=-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}+\ln \mathrm{A} \] where \(-\frac{E_a}{R}\) is the slope and \(\ln A\) is the \(y\)-intercept.…
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