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NVAMediumJEE 2024Integrated Rate Laws

JEE Chemistry 2024 Question with Solution

Consider the following first-order gas-phase reaction at constant temperature: A(g)→2B(g)+C(g)\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)} If the total pressure of the gases is found to be 200 Torr200 \, \text{Torr} after 23 s23 \, \text{s}, and 300 Torr300 \, \text{Torr} upon the complete decomposition of AA after a very long time, then the rate constant of the given reaction is _____ × 10−2 s−110^{-2} \, \text{s}^{-1} (nearest integer).

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: The first-order gas-phase reaction is

A(g)→2B(g)+C(g)\text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)}

The total pressure after 23 s23 \, \text{s} is 200 Torr200 \, \text{Torr} and the total pressure at complete decomposition is 300 Torr300 \, \text{Torr}.

Find: The value of the rate constant in the form _____ × 10−2 s−110^{-2} \, \text{s}^{-1}.

Let the initial pressure of AA be P0P_0. Since one mole of AA gives a total of three moles of gaseous products, the total pressure at completion becomes 3P03P_0.

So,

P∞=3P0=300P_{\infty} = 3P_0 = 300

Hence,

P0=100 TorrP_0 = 100 \, \text{Torr}

For a first-order reaction in terms of pressure,

K=1tln⁡P∞−P0P∞−PtK = \frac{1}{t} \ln \frac{P_{\infty} - P_0}{P_{\infty} - P_t}

Substituting the given values,

K=123ln⁡300−100300−200K = \frac{1}{23} \ln \frac{300 - 100}{300 - 200} K=2.323log⁡200100K = \frac{2.3}{23} \log \frac{200}{100} K=2.323×0.301K = \frac{2.3}{23} \times 0.301

Therefore,

K=0.0301=3.01×10−2 s−1K = 0.0301 = 3.01 \times 10^{-2} \, \text{s}^{-1}

So the nearest integer is 3.

Pressure Relation Method

Given: At time t=23 st = 23 \, \text{s}, the total pressure is Pt=200 TorrP_t = 200 \, \text{Torr}. At complete decomposition, P∞=300 TorrP_{\infty} = 300 \, \text{Torr}.

Find: The numerical value multiplying 10−2 s−110^{-2} \, \text{s}^{-1}.

If xx is the pressure-equivalent amount of AA decomposed, then at time tt:

  • pressure of AA = P0−xP_0 - x
  • pressure of products formed = 3x3x

Hence total pressure,

Pt=(P0−x)+3x=P0+2xP_t = (P_0 - x) + 3x = P_0 + 2x

Given,

Pt=200P_t = 200

Also at completion,

P∞=3P0=300⇒P0=100P_{\infty} = 3P_0 = 300 \Rightarrow P_0 = 100

Now use the integrated first-order expression in pressure form:

K=2.303tlog⁡(P∞−P0P∞−Pt)K = \frac{2.303}{t} \log \left( \frac{P_{\infty} - P_0}{P_{\infty} - P_t} \right)

Substitute:

K=2.30323log⁡(300−100300−200)K = \frac{2.303}{23} \log \left( \frac{300 - 100}{300 - 200} \right) K=2.30323log⁡(2)K = \frac{2.303}{23} \log (2)

Using log⁡2=0.301\log 2 = 0.301,

K=2.30323×0.301≈0.0303 s−1K = \frac{2.303}{23} \times 0.301 \approx 0.0303 \, \text{s}^{-1}

Thus,

K≈3×10−2 s−1K \approx 3 \times 10^{-2} \, \text{s}^{-1}

Therefore, the required answer is 3.

Common mistakes

  • Taking P∞=P0P_{\infty} = P_0 is incorrect because the number of moles of gas changes during the reaction. Here 11 mole of AA produces 33 moles of gaseous products, so the total pressure increases at constant temperature.

  • Using the first-order formula directly with total pressure PtP_t as if it were the pressure of reactant AA is wrong. The correct method is to convert the total-pressure data using P∞P_{\infty}, P0P_0, and PtP_t.

  • Missing the relation P∞=3P0P_{\infty} = 3P_0 leads to a wrong initial pressure. First determine P0=100 TorrP_0 = 100 \, \text{Torr}, then substitute into the kinetic expression.

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