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.

Integrated rate law equation for a first order gas phase reaction is given by (where Pi is initial… (JEE Main 2024) | Canvas Classes