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Chemical Kinetics and Nuclear Chemistry question

2022 · 25 Jun · Shift 1 · Q21
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Chemical Kinetics and Nuclear Chemistry question

2022 · 25 Jun · Shift 1 · Q21

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For a given chemical reaction γ\gammaγ 1A + γ\gammaγ 2B →γ\to\gamma→γ 3C + γ\gammaγ 4D Concentration of C changes from 10 mmol dm −-− 3 to 20 mmol dm −-− 3 in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm −-− 3 s −-− 1. Therefore, the rate of reaction is ‾\underline{\hspace{2cm}}​ mmol dm −-− 3 s −-− 1. (Nearest Integer)
Numerical answer
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Correct answer: 1

  1. Write the stoichiometric rate relation

For the reaction

γ1A+γ2B→γ3C+γ4D\gamma_1 A + \gamma_2 B \to \gamma_3 C + \gamma_4 Dγ1​A+γ2​B→γ3​C+γ4​D

the rate of reaction is

r=−1γ1d[A]dt=−1γ2d[B]dt=1γ3d[C]dt=1γ4d[D]dtr = -\frac{1}{\gamma_1}\frac{d[A]}{dt} = -\frac{1}{\gamma_2}\frac{d[B]}{dt} = \frac{1}{\gamma_3}\frac{d[C]}{dt} = \frac{1}{\gamma_4}\frac{d[D]}{dt}r=−γ1​1​dtd[A]​=−γ2​1​dtd[B]​=γ3​1​dtd[C]​=γ4​1​dtd[D]​
  1. Use the given change in concentration of CCC

Concentration of CCC changes from 101010 to 20 mmol dm−320\ \text{mmol dm}^{-3}20 mmol dm−3 in 10 s10\ \text{s}10 s.

So, rate of appearance of CCC is

d[C]dt=20−1010=1 mmol dm−3s−1\frac{d[C]}{dt} = \frac{20-10}{10} = 1\ \text{mmol dm}^{-3}\text{s}^{-1}dtd[C]​=1020−10​=1 mmol dm−3s−1
  1. Use the relation among species rates

Given:

  • rate of appearance of DDD is 1.51.51.5 times rate of disappearance of BBB
  • rate of disappearance of BBB is twice rate of disappearance of AAA
  • rate of appearance of DDD is experimentally 9 mmol dm−3s−19\ \text{mmol dm}^{-3}\text{s}^{-1}9 mmol dm−3s−1

Thus,

rate of appearance of D=9\text{rate of appearance of }D = 9rate of appearance of D=9

Then,

9=1.5×rate of disappearance of B9 = 1.5 \times \text{rate of disappearance of }B9=1.5×rate of disappearance of B

so

rate of disappearance of B=91.5=6\text{rate of disappearance of }B = \frac{9}{1.5} = 6rate of disappearance of B=1.59​=6

Also,

rate of disappearance of B=2×rate of disappearance of A\text{rate of disappearance of }B = 2 \times \text{rate of disappearance of }Arate of disappearance of B=2×rate of disappearance of A

therefore

rate of disappearance of A=62=3\text{rate of disappearance of }A = \frac{6}{2} = 3rate of disappearance of A=26​=3

So the species rates are:

−d[A]dt=3,−d[B]dt=6,d[C]dt=1,d[D]dt=9-\frac{d[A]}{dt}=3, \qquad -\frac{d[B]}{dt}=6, \qquad \frac{d[C]}{dt}=1, \qquad \frac{d[D]}{dt}=9−dtd[A]​=3,−dtd[B]​=6,dtd[C]​=1,dtd[D]​=9
  1. Determine stoichiometric coefficients ratio

Since

−d[A]/dtγ1=−d[B]/dtγ2=d[C]/dtγ3=d[D]/dtγ4=r\frac{-d[A]/dt}{\gamma_1} = \frac{-d[B]/dt}{\gamma_2} = \frac{d[C]/dt}{\gamma_3} = \frac{d[D]/dt}{\gamma_4} = rγ1​−d[A]/dt​=γ2​−d[B]/dt​=γ3​d[C]/dt​=γ4​d[D]/dt​=r

we get

γ1:γ2:γ3:γ4=3:6:1:9\gamma_1 : \gamma_2 : \gamma_3 : \gamma_4 = 3 : 6 : 1 : 9γ1​:γ2​:γ3​:γ4​=3:6:1:9

Hence the smallest integer stoichiometric coefficients are

γ1=3,γ2=6,γ3=1,γ4=9\gamma_1=3,\quad \gamma_2=6,\quad \gamma_3=1,\quad \gamma_4=9γ1​=3,γ2​=6,γ3​=1,γ4​=9
  1. Calculate the rate of reaction

Using species CCC:

r=1γ3d[C]dt=11×1=1 mmol dm−3s−1r = \frac{1}{\gamma_3}\frac{d[C]}{dt} = \frac{1}{1}\times 1 = 1\ \text{mmol dm}^{-3}\text{s}^{-1}r=γ3​1​dtd[C]​=11​×1=1 mmol dm−3s−1

Alternatively using DDD:

r=19×9=1 mmol dm−3s−1r = \frac{1}{9}\times 9 = 1\ \text{mmol dm}^{-3}\text{s}^{-1}r=91​×9=1 mmol dm−3s−1

So the rate of reaction is

1 mmol dm−3s−1\boxed{1\ \text{mmol dm}^{-3}\text{s}^{-1}}1 mmol dm−3s−1​
  1. Comparison with stored answer

Stored correct answer = 111

Our derived answer also equals 111, so it agrees.

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