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JEE Advanced 2021 Paper 2 · Chemistry · Chemical Kinetics · medium

JEE Advanced 2021 Paper 2 Chemistry: Paragraph 3

The gas-phase decomposition A(g)→2B(g)+C(g)\mathrm{A(g)\to 2B(g)+C(g)} is first order in A\mathrm{A}. Pure A\mathrm{A} is introduced into a rigid vessel at 400 K400\ \mathrm{K} at an initial pressure of 0.500 atm0.500\ \mathrm{atm}, and the total pressure is measured as the reaction proceeds. After 20.020.0 minutes the total pressure is 0.800 atm0.800\ \mathrm{atm}. Assume ideal gas behaviour, that B\mathrm{B} and C\mathrm{C} do not react further, and constant temperature throughout.

Question 1

Numerical answer

The first-order rate constant kk at 400 K400\ \mathrm{K}, in units of 10−2 min−110^{-2}\ \mathrm{min^{-1}}, is ______. Give the answer correct to two decimal places.
Show answer and solution

Answer: 1.78

Let xx be the pressure of A\mathrm{A} consumed. Each mole of A gives 33 moles of gas, so the total pressure is Pt=(P0−x)+2x+x=P0+2xP_t=(P_0-x)+2x+x=P_0+2x. At 20.020.0 min, 0.800=0.500+2x0.800=0.500+2x, giving x=0.150x=0.150 atm, so PA=0.500−0.150=0.350P_A=0.500-0.150=0.350 atm. For a first-order reaction k=1tln⁡P0PA=120.0ln⁡0.5000.350=0.356720.0=1.78×10−2 min−1k=\dfrac{1}{t}\ln\dfrac{P_0}{P_A}=\dfrac{1}{20.0}\ln\dfrac{0.500}{0.350}=\dfrac{0.3567}{20.0}=1.78\times10^{-2}\ \mathrm{min^{-1}}. The key step is relating total pressure to the partial pressure of A through the stoichiometry; using PtP_t directly in the logarithm would be wrong.

Question 2

Numerical answer

The time, in minutes, at which the total pressure reaches 1.100 atm1.100\ \mathrm{atm} at 400 K400\ \mathrm{K} is ______. Give the answer correct to two decimal places.
Show answer and solution

Answer: 51.38

Using Pt=P0+2xP_t=P_0+2x with Pt=1.100P_t=1.100 atm gives x=0.300x=0.300 atm, so PA=0.500−0.300=0.200P_A=0.500-0.300=0.200 atm. For first-order kinetics t=1kln⁡P0PAt=\dfrac{1}{k}\ln\dfrac{P_0}{P_A}, with k=120.0ln⁡0.5000.350=1.7834×10−2 min−1k=\dfrac{1}{20.0}\ln\dfrac{0.500}{0.350}=1.7834\times10^{-2}\ \mathrm{min^{-1}} retained without rounding. Then t=ln⁡(0.500/0.200)1.7834×10−2=0.91630.017834=51.38t=\dfrac{\ln(0.500/0.200)}{1.7834\times10^{-2}}=\dfrac{0.9163}{0.017834}=51.38 min. As a check, the consistency with the earlier data follows from the ratio t=20.0×ln⁡2.5/ln⁡(0.5/0.35)=20.0×0.9163/0.3567=51.38t=20.0\times\ln2.5/\ln(0.5/0.35)=20.0\times0.9163/0.3567=51.38 minutes. The reaction is thus about 2.52.5 times the earlier time because the pressure of A must fall to 40%40\% of its initial value.
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