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MCQMediumJEE 2024Integrated Rate Laws

JEE Chemistry 2024 Question with Solution

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):

  • A

    k=2.303t×log⁡(Pi2Pi−Pt)k = \frac{2.303}{t} \times \log \left(\frac{P_i}{2P_i - P_t}\right)

  • B

    k=2.303t×log⁡(2Pi2Pi−Pt)k = \frac{2.303}{t} \times \log \left(\frac{2P_i}{2P_i - P_t}\right)

  • C

    k=2.303t×log⁡(2Pi−PtPi)k = \frac{2.303}{t} \times \log \left(\frac{2P_i - P_t}{P_i}\right)

  • D

    k=2.303t×(Pi2Pi−Pt)k = \frac{2.303}{t} \times \left(\frac{P_i}{2P_i - P_t}\right)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A first order gas phase reaction is expressed in terms of initial pressure PiP_i and total pressure at time tt as PtP_t.

Find: The correct integrated rate law expression.

For a first-order reaction, the integrated rate law is

k=2.303tlog⁡[A]0[A]k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}

In the gas phase, concentration is proportional to pressure. Therefore,

[A]0→Pi[A]_0 \rightarrow P_i

and the pressure of reactant at time tt is obtained from the pressure relation.

For the reaction

A→B+CA \rightarrow B + C

let the decrease in pressure of AA be xx. Then,

Initial pressures: Pi, 0, 0\text{Initial pressures: } P_i,\ 0,\ 0 At time t:Pi−x, x, x\text{At time } t: P_i - x,\ x,\ x

So the total pressure at time tt is

Pt=(Pi−x)+x+x=Pi+xP_t = (P_i - x) + x + x = P_i + x

Hence,

x=Pt−Pix = P_t - P_i

Therefore, the partial pressure of AA at time tt is

PA=Pi−x=Pi−(Pt−Pi)=2Pi−PtP_A = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t

Substituting into the first-order integrated rate law,

k=2.303tlog⁡Pi2Pi−Ptk = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}

Therefore, the correct option is A.

Using pressure relation explicitly

Given: Initial pressure is PiP_i and total pressure after time tt is PtP_t.

Find: The integrated rate law in pressure form.

For a first-order reaction,

k=2.303tlog⁡initial pressurepressure at time tk = \frac{2.303}{t} \log \frac{\text{initial pressure}}{\text{pressure at time } t}

From the reaction stoichiometry,

A→B+CA \rightarrow B + C

if xx pressure of AA reacts, then total pressure increases by xx because one mole of reactant forms two moles of products.

Thus,

Pt=Pi+xP_t = P_i + x

so the remaining pressure of reactant is

Pi−x=Pi−(Pt−Pi)=2Pi−PtP_i - x = P_i - (P_t - P_i) = 2P_i - P_t

Now substitute in the integrated law:

k=2.303tlog⁡Pi2Pi−Ptk = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t}

This matches Option 1, so the correct answer is A.

The solution notes that Option 2 and Option 4 are the same as Option 1, but from the listed options they are not identical. The derived expression clearly matches Option 1 only.

Common mistakes

  • Using PtP_t directly in place of the reactant pressure at time tt is incorrect because PtP_t is the total pressure, not the partial pressure of reactant A. First find the remaining reactant pressure as 2Pi−Pt2P_i - P_t.

  • Reversing the logarithm to write log⁡(2Pi−PtPi)\log \left(\frac{2P_i - P_t}{P_i}\right) gives the negative of the required expression. For a first-order decay form, use log⁡(Pi2Pi−Pt)\log \left(\frac{P_i}{2P_i - P_t}\right).

  • Ignoring the stoichiometry of A→B+CA \rightarrow B + C leads to a wrong pressure relation. The increase in total pressure comes from formation of two product species, so write the pressure table before substituting into the rate law.

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