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JEE Advanced 2024 Paper 1 · Chemistry · Chemical Kinetics · medium

The gas-phase decomposition A(g)→ 2 B(g)+C(g) is first order in A and is carried out in a rigid vessel at constant temperature, starting with pure A…

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The gas-phase decomposition A(g)→2 B(g)+C(g)\mathrm{A(g)}\rightarrow 2\,\mathrm{B(g)}+\mathrm{C(g)} is first order in A and is carried out in a rigid vessel at constant temperature, starting with pure A at 100 kPa100\ \mathrm{kPa}. After 4040 minutes the total pressure is 250 kPa250\ \mathrm{kPa}. The total pressure after 6060 minutes is
  1. A. 262.5 kPa262.5\ \mathrm{kPa}
  2. B. 287.5 kPa287.5\ \mathrm{kPa}
  3. C. 268.75 kPa268.75\ \mathrm{kPa}
  4. D. 275 kPa275\ \mathrm{kPa}
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Answer: (D) 275 kPa275\ \mathrm{kPa}

If pAp_A is the partial pressure of A left, then pB=2(P0−pA)p_B=2(P_0-p_A) and pC=P0−pAp_C=P_0-p_A, so Ptotal=pA+3(P0−pA)=3P0−2pAP_{total}=p_A+3(P_0-p_A)=3P_0-2p_A. At 4040 min: 250=300−2pA⇒pA=25250=300-2p_A\Rightarrow p_A=25 kPa, so pA/P0=1/4p_A/P_0=1/4, i.e. two half-lives, giving t1/2=20t_{1/2}=20 min. At 6060 min (three half-lives): pA=100/8=12.5p_A=100/8=12.5 kPa, and Ptotal=300−2(12.5)=275P_{total}=300-2(12.5)=275 kPa. The other values arise from using P=2P0−pAP=2P_0-p_A-type relations, from assuming a different number of half-lives, or from forgetting the factor 2 on pAp_A in the total pressure expression.
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