JEE Main · 2024 · Shift-ImediumCK-053

Integrated rate law equation for a first order gas phase reaction is given by (where Pi is initial pressure and Pt is…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

Integrated rate law equation for a first order gas phase reaction is given by (where PiP_i is initial pressure and PtP_t is total pressure at time tt):

Options
  1. a

    k=2.303t×logPi(2PiPt)k = \frac{2.303}{t} \times \log\frac{P_i}{(2P_i - P_t)}

  2. b

    k=2.303t×log2Pi(2PiPt)k = \frac{2.303}{t} \times \log\frac{2P_i}{(2P_i - P_t)}

  3. c

    k=2.303t×log(2PiPt)Pik = \frac{2.303}{t} \times \log\frac{(2P_i - P_t)}{P_i}

  4. d

    k=2.303t×Pi(2PiPt)k = \frac{2.303}{t} \times \frac{P_i}{(2P_i - P_t)}

Correct Answera

k=2.303t×logPi(2PiPt)k = \frac{2.303}{t} \times \log\frac{P_i}{(2P_i - P_t)}

Detailed Solution

For a first order gas phase reaction A(g)B(g)+C(g)\mathrm{A(g) \rightarrow B(g) + C(g)}:

Let initial pressure of A = PiP_i, decrease in pressure = pp.

At time tt: PA=PipP_A = P_i - p, total pressure Pt=(Pip)+p+p=Pi+pP_t = (P_i - p) + p + p = P_i + p

So p=PtPip = P_t - P_i, and PA=Pi(PtPi)=2PiPtP_A = P_i - (P_t - P_i) = 2P_i - P_t

For first order: k=2.303tlogPiPA=2.303tlogPi2PiPtk = \frac{2.303}{t}\log\frac{P_i}{P_A} = \frac{2.303}{t}\log\frac{P_i}{2P_i - P_t}

Answer: Option (1)

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Chemical Kinetics) inside The Crucible, our adaptive practice platform.