JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
The first order rate constant for the decomposition of \(\mathrm{CaCO}_{3}\) at \(700\, \mathrm{~K}\) is \(6.36 \times 10^{-3}\, \mathrm{~s}^{-1}\) and activation energy is \(209\, \mathrm{~kJ} \,\mathrm{~mol}^{-1}\). Its rate constant (in \(\mathrm{s}^{-1}\) ) at \(500\, \mathrm{~K}\) is \(\mathrm{x} \times 10^{-6}\). The value of \(\mathrm{x}\) is ..... . Nearest integer) Given \(\mathrm{R}=8.31\, \mathrm{~J} \,\mathrm{~K}^{-1} \,\mathrm{~mol}^{-1} ; \log 6.36 \times 10^{-3}=-2.19\) \(\left.10^{-4.79}=1.62 \times 10^{-5}\right]\)
- A \(16\)
- B \(1.6\)
- C \(0.16\)
- D \(160\)
Answer & Solution
Correct Answer
(A) \(16\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{700}=6.36 \times 10^{-3} \mathrm{~s}^{-1}\) \(\mathrm{~K}_{600}=x \times 10^{-6} \mathrm{~s}^{-1}\) \(\mathrm{E}_{\mathrm{a}}=209 \mathrm{~kJ} / \mathrm{mol}\) Applying ;…
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