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First Order Reaction

For a first o...

Question

For a first order gas phase reaction:

$A_{(g)} \to 2 B_{(g)} + C_{(g)}$ .

$P_{0}$ be initial pressure of $A$ and $P$ , the total pressure at time ' $t^{'}$ . Integrated rate equation is:

\frac{2.303}{t} log (\frac{P_{0}}{P_{0} - P_{t}})

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\frac{2.303}{t} log (\frac{2 P_{0}}{3 P_{0} - P_{t}})

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\frac{2.303}{t} log (\frac{P_{0}}{2 P_{0} - P_{t}})

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\frac{2.303}{t} log (\frac{2 P_{0}}{2 P_{0} - P_{t}})

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Solution

The correct option is B $\frac{2.303}{t} log (\frac{2 P_{0}}{3 P_{0} - P_{t}})$
The given reaction is :

$A_{(g)} \to 2 B_{(g)} + C_{(g)}$

$A$ $B$ $C$
Initial $P_{0}$ $0$ $0$
Final $P_{0} - P$
$2 P$ $P$
Total pressure at time $(t) = P_{0} - P + 2 P + P = P_{t}$

$\Rightarrow P_{t} = P_{0} + 2 P$

$P_{t} - P_{0} = 2 P \Rightarrow P = \frac{P_{t} - P_{0}}{2}$

$k = \frac{2.303}{t} log [\frac{P_{0}}{P_{0} - P}] = \frac{2.303}{t} log ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{P_{0}}{P_{0} - (\frac{P_{t} - P_{0}}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$= \frac{2.303}{t} log (\frac{2 P_{0}}{2 P_{0} - P_{t} + P_{0}}) = \frac{2.303}{t} log (\frac{2 P_{0}}{3 P_{0} - P_{t}})$

Hence, option $B$ is correct.

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For a first order gas phase reaction: A(g)→2B(g)+C(g). P0 be initial pressure of A and P, the total pressure at time 't′. Integrated rate equation is:

For a first order gas phase reaction:

$A_{(g)} \to 2 B_{(g)} + C_{(g)}$ .

$P_{0}$ be initial pressure of $A$ and $P$ , the total pressure at time ' $t^{'}$ . Integrated rate equation is: