Chemical Kinetics and Nuclear ChemistryHard

Question

For the first-order reaction A(g) → 2B(g) + C(g), the total pressure after time t from the start of reaction with A is P and after infinite time, it is P. Then the rate constant of the reaction is

Options

A.$\frac{1}{t}\ln\frac{P_{\infty}}{P}$
B.$\frac{1}{t}.\ln\frac{2P_{\infty}}{3\left( P_{\infty} - P \right)}$
C.$\frac{1}{t}.\ln\frac{2P_{\infty}}{3P_{\infty} - P}$
D.$\frac{1}{t}.\ln\frac{2P_{\infty}}{P_{\infty} - 3P}$

Solution

A ? 2B + C

t = 0 P0 0 0

t = t P0 – x 2x x

t = ∞ ≈ 0 2P0 P0

From question: $P_{\infty} = 2P_{0} + P_{0}$

$\Rightarrow P_{0} = \frac{P_{\infty}}{3}\text{ and }P = \left( P_{0} - x \right) + 2x + x $$$\Rightarrow x = \frac{P - P_{0}}{2} = \frac{3P - P_{\infty}}{6}$$

$\text{Now, }K = \frac{1}{t}.\ln\frac{P_{A}^{o}}{P_{A}} = \frac{1}{t}.\ln\frac{P_{o}}{P_{o} - x} $$${= \frac{1}{t}.\ln\frac{P_{\infty}/3}{\frac{P_{\infty}}{3} - \frac{3P - P_{\infty}}{6}} }{= \frac{1}{t}.\ln\frac{2P_{\infty}}{3\left( P_{\infty} - P \right)}}$$

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