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AIIMS 2011
Exam
AIIMS
Year
2011
Subject
Chemistry
Chapter
Chemical Kinetics

AIIMS 2011 · Chemistry · Chemical Kinetics

For a first order gas phase reaction- A_ (g) arrow 2 B_ (g) +C_ (g)...

For a first order gas phase reaction- A_{(g)} \rightarrow 2 B_{(g)}+C_{(g)} P0P_0 be initial pressure of AA and PtP_t the total pressure at time ' tt '. Integrated rate equation is

Options

  1. A2.303tlog⁡(P0P0−Pt)\frac{2.303}{t} \log \left(\frac{P_0}{P_0-P_t}\right)
  2. B2.303tlog⁡(2P03P0−Pt)\frac{2.303}{t} \log \left(\frac{2 P_0}{3 P_0-P_t}\right)
  3. C2.303tlog⁡(P02P0−Pt)\frac{2.303}{t} \log \left(\frac{P_0}{2 P_0-P_t}\right)
  4. D2.303tlog⁡(2P02P0−Pt)\frac{2.303}{t} \log \left(\frac{2 P_0}{2 P_0-P_t}\right)

Correct answer

B. 2.303tlog⁡(2P03P0−Pt)\frac{2.303}{t} \log \left(\frac{2 P_0}{3 P_0-P_t}\right)

Step-by-step solution

\begin{array}{lccc} \text{ } & \text{ } & A_{(g)} & \longrightarrow 2 B_{(g)}+C_{(g)} \\ \text{ Initial }: & P_0 & 0 & 0 \\ & P_0-P & 2 P & P \end{array} Total pressure at time (t)=P0−P+2P+P=Pt(t)=P_0-P+2 P+P=P_t \begin{aligned} \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 \left[\frac{P_0}{P_0-P}\right]=\frac{2.303}{t} \log \left[\frac{P_0}{P_0-\left(\frac{P_t-P_0}{2}\right)}\right] \\ = & \frac{2.303}{t} \log \left(\frac{2 P_0}{2 P_0-P_t+P_0}\right)=\frac{2.303}{t} \log \left(\frac{2 P_0}{3 P_0-P_t}\right) \end{aligned}
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