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COMEDK · Chemistry · 19. Chemical Kinetics

Given below a first order reaction in the gas phase
\(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})\)
If the initial pressure of the system is \(\mathrm{P}_{\mathrm{i}}\) and the total pressure at \(\mathrm{t}\) seconds is \(\mathrm{P}_{\mathrm{t}}\), the rate constant \(\mathrm{k}\) for the reaction is:

  1. A \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}+\mathrm{x}\right)}\)
  2. B \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_{\mathrm{t}}}\)
  3. C \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{P}_{\mathrm{i}}}{\left(\mathrm{P}_{\mathrm{i}}+\mathrm{x}\right)}\)
  4. D \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{k}=\dfrac{2.303}{\mathrm{t}} \log \dfrac{\mathrm{P}_{\mathrm{i}}}{\left(2 \mathrm{P}_{\mathrm{i}}-\mathrm{P}_{\mathrm{t}}\right)}\)

Step-by-step Solution

Detailed explanation

The reaction is \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{B}(\mathrm{g}) + \mathrm{C}(\mathrm{g})\). At \(t = 0\), the pressure of A is \(P_i\), and the pressures of B and C are \(0\). At time \(t\), let the decrease in pressure of A be \(x\). Then the pressures of B and C…