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Chemical Kinetics and Nuclear Chemistry question

2023 · 11 Apr · Shift 2 · Q13
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Chemical Kinetics and Nuclear Chemistry question

2023 · 11 Apr · Shift 2 · Q13

JEE MainChemistryChemical Kinetics and Nuclear ChemistryMCQ+4 / −1

For a chemical reaction A+B→\mathrm{A}+\mathrm{B} \rightarrowA+B→ Product, the order is 1 with respect to A\mathrm{A}A and B\mathrm{B}B.

Rate\mathrm{Rate}Rate
mol L−1 S−1\mathrm{mol~L^{-1}~S^{-1}}mol L−1 S−1		 [A]\mathrm{[A]}[A]
mol L−1\mathrm{mol~L^{-1}}mol L−1		 [B]\mathrm{[B]}[B]
mol L−1\mathrm{mol~L^{-1}}mol L−1
0.10 20 0.5
0.40 xxx 0.5
0.80 40 yyy

What is the value of xxx and yyy ?

  1. A
    80 and 4
  2. B
    160 and 4
  3. C
    80 and 2
  4. D
    40 and 4
View written solutionFree

Correct answer: C

  1. Write the rate law

Since the reaction is first order with respect to both AAA and BBB,

Rate=k[A][B]\text{Rate} = k[A][B]Rate=k[A][B]

  1. Use the first experiment to find kkk

Given:

Rate=0.10,[A]=20,[B]=0.5\text{Rate} = 0.10, \quad [A]=20, \quad [B]=0.5Rate=0.10,[A]=20,[B]=0.5

So,

0.10=k(20)(0.5)0.10 = k(20)(0.5)0.10=k(20)(0.5)

0.10=10k0.10 = 10k0.10=10k

k=0.01k = 0.01k=0.01

  1. Use the second experiment to find xxx

Given:

Rate=0.40,[A]=x,[B]=0.5\text{Rate} = 0.40, \quad [A]=x, \quad [B]=0.5Rate=0.40,[A]=x,[B]=0.5

Using k=0.01k=0.01k=0.01,

0.40=0.01(x)(0.5)0.40 = 0.01(x)(0.5)0.40=0.01(x)(0.5)

0.40=0.005x0.40 = 0.005x0.40=0.005x

x=0.400.005=80x = \frac{0.40}{0.005} = 80x=0.0050.40​=80

  1. Use the third experiment to find yyy

Given:

Rate=0.80,[A]=40,[B]=y\text{Rate} = 0.80, \quad [A]=40, \quad [B]=yRate=0.80,[A]=40,[B]=y

Again,

0.80=0.01(40)(y)0.80 = 0.01(40)(y)0.80=0.01(40)(y)

0.80=0.4y0.80 = 0.4y0.80=0.4y

y=0.800.4=2y = \frac{0.80}{0.4} = 2y=0.40.80​=2

  1. Final answer

x=80,y=2x = 80, \quad y = 2x=80,y=2

So the correct option is C.

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