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QJEEAdv21Ch4ac0: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Ch4ac0 Solution: For the following reaction, 2 X + Y ⟶ k P the rate of reaction…

For the following reaction,
2X+Y⟶kP the rate of reaction is d[P]dt=k[X]{{d[P]} \over {dt}} = k[X] . Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are)

(Use : ln 2 = 0.693)

Multiple Choice Options

  • A The rate constant, k, of the reaction is 13.86 ×\times 10 −- 4 s −- 1
  • B Half-life of X is 50 s.
  • C At 50 s, −- d[X]dt{{d[X]} \over {dt}} = 13.86 ×\times 10 −- 3 mol L −- 1 s −- 1.
  • D At 100 s, −- d[Y]dt{{d[Y]} \over {dt}} = 3.46 ×\times 10 −- 3 mol L −- 1 s −- 1.

Step-by-Step Verified Solution

JEE Advanced 2021 Paper 2 Online Chemistry - Chemical Kinetics and Nuclear Chemistry Question 13 English Explanation
As the concentration of reactant becomes half at t = 50 s. So, half-time of reaction is 50 s.

Given,

dPdt=k[X]1{{dP} \over {dt}} = k{[X]^1}

⇒−12dXdt=dPdt=k[X]1\Rightarrow - {1 \over 2}{{dX} \over {dt}} = {{dP} \over {dt}} = k{[X]^1}

−dXdt=2k[X]1- {{dX} \over {dt}} = 2k{[X]^1}

−dX[X]=2k dt{{ - dX} \over {[X]}} = 2k\,dt

−∫X0XdXX=2k ∫0tdt- \int_{{X_0}}^X {{{dX} \over X} = 2k\,\int_0^t {dt} }

−[ln⁡X]X0X=2k[t]0t- [\ln X]_{{X_0}}^X = 2k[t]_0^t

−ln⁡X+ln⁡X0=2kt- \ln X + \ln {X_0} = 2kt

ln⁡X0X=2kt\ln {{{X_0}} \over X} = 2kt

At t1/2, X=X02X = {{{X_0}} \over 2}

ln⁡2X0X0=2kt1/2\ln {{2{X_0}} \over {{X_0}}} = 2k{t_{1/2}}

k=ln⁡22t1/2=0.693100=6.93×10−3s−1k = {{\ln 2} \over {2{t_{1/2}}}} = {{0.693} \over {100}} = 6.93 \times {10^{ - 3}}{s^{ - 1}}

At t = 50 s, −dxdt=2k[X]1;[X]=1- {{dx} \over {dt}} = 2k{[X]^1};[X] = 1 mol

So, −dXdt=2×6.93×10−3×1=13.86×10−3{{ - dX} \over {dt}} = 2 \times 6.93 \times {10^{ - 3}} \times 1 = 13.86 \times {10^{ - 3}} mol L-1 s-1

At t = 100 s, −12dXdt=−dYdt⇒−dYdt=k[Y]1;[Y]=12- {1 \over 2}{{dX} \over {dt}} = {{ - dY} \over {dt}} \Rightarrow {{ - dY} \over {dt}} = k{[Y]^1};[Y] = {1 \over 2}

−dYdt=6.93×10−3×12=3.46×10−3- {{dY} \over {dt}} = 6.93 \times {10^{ - 3}} \times {1 \over 2} = 3.46 \times {10^{ - 3}} mol L-1 s-1

So, options (b), (c) and (d) are correct.

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