MCQMediumJEE 2025Integrated Rate Laws

JEE Chemistry 2025 Question with Solution

A(g) \rightarrow B(g) + C(g) is a first order reaction.

Table showing system pressure at time t as P_t and at infinite time as P_\infty for the gas reaction.

The reaction was started with reactant A only. Which of the following expression is correct for rate constant kk ?

  • A

    k=1tln2(PPt)Ptk = \frac{1}{t} \ln \frac{2(P_\infty - P_t)}{P_t}

  • B

    k=1tlnPPtk = \frac{1}{t} \ln \frac{P_\infty}{P_t}

  • C

    k=1tlnP2(PPt)k = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)}

  • D

    k=1tlnP(PPt)k = \frac{1}{t} \ln \frac{P_\infty}{(P_\infty - P_t)}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A(g)B(g)+C(g)A(g) \rightarrow B(g) + C(g) is a first-order reaction. The total pressure at time tt is PtP_t and at completion it is PP_\infty.

Find: The correct expression for the rate constant kk.

Let the initial pressure of AA be P0P_0. If pressure corresponding to reacted AA at time tt is pp, then

PA=P0p,PB=p,PC=pP_A = P_0 - p, \qquad P_B = p, \qquad P_C = p

So the total pressure at time tt is

Pt=(P0p)+p+p=P0+pP_t = (P_0 - p) + p + p = P_0 + p

Hence,

p=PtP0p = P_t - P_0

and therefore

PA=P0(PtP0)=2P0PtP_A = P_0 - (P_t - P_0) = 2P_0 - P_t

At t=t = \infty, the reaction is complete, so

P=2P0P_\infty = 2P_0

Thus,

P0=P2P_0 = \frac{P_\infty}{2}

Substituting into PAP_A,

PA=2(P2)Pt=PPtP_A = 2\left(\frac{P_\infty}{2}\right) - P_t = P_\infty - P_t

For a first-order reaction,

k=1tlnP0PAk = \frac{1}{t} \ln \frac{P_0}{P_A}

Now substitute P0=P2P_0 = \frac{P_\infty}{2} and PA=PPtP_A = P_\infty - P_t:

k=1tlnP/2PPtk = \frac{1}{t} \ln \frac{P_\infty/2}{P_\infty - P_t} k=1tlnP2(PPt)k = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)}

Therefore, the correct option is C.

Pressure-Stoichiometry Approach

Given: Only reactant AA is present initially for the reaction A(g)B(g)+C(g)A(g) \rightarrow B(g) + C(g).

Find: Expression of kk in terms of PtP_t and PP_\infty.

The key idea is to relate the partial pressure of unreacted AA to the measured total pressure.

At any time tt, if reacted pressure is pp, then the species pressures are

PA=P0p,PB=p,PC=pP_A = P_0 - p, \qquad P_B = p, \qquad P_C = p

So,

Pt=PA+PB+PC=(P0p)+p+p=P0+pP_t = P_A + P_B + P_C = (P_0 - p) + p + p = P_0 + p

which gives

p=PtP0p = P_t - P_0

Hence,

PA=P0p=P0(PtP0)=2P0PtP_A = P_0 - p = P_0 - (P_t - P_0) = 2P_0 - P_t

At completion,

PA()=0,PB()=P0,PC()=P0P_A(\infty) = 0, \qquad P_B(\infty) = P_0, \qquad P_C(\infty) = P_0

Therefore,

P=0+P0+P0=2P0P_\infty = 0 + P_0 + P_0 = 2P_0

So,

P0=P2P_0 = \frac{P_\infty}{2}

and

PA=PPtP_A = P_\infty - P_t

Using the integrated first-order law,

k=1tln[A]0[A]k = \frac{1}{t} \ln \frac{[A]_0}{[A]}

For gases at constant temperature, concentration is proportional to partial pressure, so

k=1tlnP0PAk = \frac{1}{t} \ln \frac{P_0}{P_A}

Substituting the pressure relations,

k=1tlnP/2PPt=1tlnP2(PPt)k = \frac{1}{t} \ln \frac{P_\infty/2}{P_\infty - P_t} = \frac{1}{t} \ln \frac{P_\infty}{2(P_\infty - P_t)}

This matches option (3), so the correct option is C.

Common mistakes

  • Using total pressure PtP_t directly in the first-order formula is incorrect because the rate law depends on the partial pressure of unreacted AA, not the total system pressure. First express PAP_A in terms of PtP_t and PP_\infty, then substitute into the integrated law.

  • Assuming PA=PPtP_A = P_\infty - P_t without deriving P=2P0P_\infty = 2P_0 can lead to confusion. The stoichiometry AB+CA \rightarrow B + C doubles the total number of gas moles at completion, so this relation must be established first.

  • Forgetting that only AA is present initially is a conceptual error. If products were initially present, the pressure relations would change. Use the given initial condition to write PB=0P_B = 0 and PC=0P_C = 0 at t=0t = 0.

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