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AIIMS 2009
Exam
AIIMS
Year
2009
Subject
Chemistry
Chapter
Chemical Kinetics

AIIMS 2009 · Chemistry · Chemical Kinetics

In a homogenous reaction A B+C+D the initial pressure was P_0 and...

In a homogenous reaction A⟶B+C+DA \longrightarrow B+C+D the initial pressure was P0P_0 and after time tt it was PP. Expression for rate constant kk in terms of P0P_0, PP and tt will beIn a homogenous reaction A B+C+D the initial pressure was P_0 and after time t it was P . Expression for rate constant

Options

  1. Ak=2.303tlog⁡2P03P0−Pk=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-P}
  2. Bk=2.303tlog⁡2P0P0−Pk=\frac{2.303}{t} \log \frac{2 P_0}{P_0-P}
  3. Ck=2.303tlog⁡3P0−P2P0k=\frac{2.303}{t} \log \frac{3 P_0-P}{2 P_0}
  4. Dk=2.303tlog⁡2P03P0−2Pk=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-2 P}

Correct answer

A. k=2.303tlog⁡2P03P0−Pk=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-P}

Step-by-step solution

A⟶B+C+DA \longrightarrow B+C+D  Initial a000 After time ta−xxxx\begin{array}{lllll}\text{ Initial } & a & 0 & 0 & 0 \\ \text{ After time } t a-x & x & x & x\end{array} In a homogenous reaction A B+C+D the initial pressure was P_0 and after time t it was P . Expression for rate constant
From (i), P0+2x=PP_0+2 x=P \quad or x=P−P02\quad x=\frac{P-P_0}{2} From rate equation k=2.303tlog⁡aa−xk=\frac{2.303}{t} \log \frac{a}{a-x} =2.303tlog⁡P0P0−(P−P02)=2.303tlog⁡2P03P0−P=\frac{2.303}{t} \log \frac{P_0}{P_0-\left(\frac{P-P_0}{2}\right)}=\frac{2.303}{t} \log \frac{2 P_0}{3 P_0-P}
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