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Chemical Kinetics and Nuclear Chemistry question

2014 · Shift 0 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2014 · Shift 0 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1
For the non – stoichiometre reaction 2A + B →\to→ C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.
Initial Concentration (A) Initial Concentration (B) Initial rate of formation of C (mol L-1 s-1)
0.1 M 0.1 M 1.2 x 10-3
0.1 M 0.2 M 1.2 x 10-3
0.2 M 0.1 M 2.4 x 10-3
The rate law for the formation of C is:
  1. A
    dcdt=k[A][B]2{{dc} \over {dt}} = k[A]{[B]^2}dtdc​=k[A][B]2
  2. B
    dcdt=k[A]{{dc} \over {dt}} = k[A]dtdc​=k[A]
  3. C
    dcdt=k[A][B]{{dc} \over {dt}} = k[A]{[B]}dtdc​=k[A][B]
  4. D
    dcdt=k[A2][B]{{dc} \over {dt}} = k[A^2]{[B]}dtdc​=k[A2][B]
View written solutionFree

Correct answer: B

  1. Assume a general rate law

For the reaction 2A+B→C+D2A + B \to C + D2A+B→C+D let the rate of formation of CCC be d[C]dt=k[A]m[B]n\frac{d[C]}{dt} = k[A]^m[B]^ndtd[C]​=k[A]m[B]n where mmm and nnn are the orders with respect to AAA and BBB.

  1. Use experiments 1 and 2 to find order in BBB

From the data:

  • Exp 1: [A]=0.1[A]=0.1[A]=0.1, [B]=0.1[B]=0.1[B]=0.1, rate =1.2×10−3=1.2\times 10^{-3}=1.2×10−3
  • Exp 2: [A]=0.1[A]=0.1[A]=0.1, [B]=0.2[B]=0.2[B]=0.2, rate =1.2×10−3=1.2\times 10^{-3}=1.2×10−3

Here, [A][A][A] is constant and [B][B][B] is doubled, but the rate remains unchanged.

So, rate2rate1=([B]2[B]1)n=(0.20.1)n=2n=1\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[B]_2}{[B]_1}\right)^n = \left(\frac{0.2}{0.1}\right)^n = 2^n = 1rate1​rate2​​=([B]1​[B]2​​)n=(0.10.2​)n=2n=1 Hence, n=0n=0n=0

So the reaction is zero order in BBB.

  1. Use experiments 1 and 3 to find order in AAA

From the data:

  • Exp 1: [A]=0.1[A]=0.1[A]=0.1, [B]=0.1[B]=0.1[B]=0.1, rate =1.2×10−3=1.2\times 10^{-3}=1.2×10−3
  • Exp 3: [A]=0.2[A]=0.2[A]=0.2, [B]=0.1[B]=0.1[B]=0.1, rate =2.4×10−3=2.4\times 10^{-3}=2.4×10−3

Here, [B][B][B] is constant and [A][A][A] is doubled; the rate also doubles.

Thus, rate3rate1=([A]3[A]1)m=(0.20.1)m=2m=2.4×10−31.2×10−3=2\frac{\text{rate}_3}{\text{rate}_1} = \left(\frac{[A]_3}{[A]_1}\right)^m = \left(\frac{0.2}{0.1}\right)^m = 2^m = \frac{2.4\times 10^{-3}}{1.2\times 10^{-3}} = 2rate1​rate3​​=([A]1​[A]3​​)m=(0.10.2​)m=2m=1.2×10−32.4×10−3​=2 Therefore, m=1m=1m=1

So the reaction is first order in AAA.

  1. Write the final rate law

Since m=1m=1m=1 and n=0n=0n=0, d[C]dt=k[A]1[B]0=k[A]\frac{d[C]}{dt} = k[A]^1[B]^0 = k[A]dtd[C]​=k[A]1[B]0=k[A]

  1. Match with options

This corresponds to:

Option B: dCdt=k[A]\dfrac{dC}{dt}=k[A]dtdC​=k[A]

  1. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.

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