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Chemical Kinetics and Nuclear Chemistry question

2010 · Shift 1 · Q12
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Chemical Kinetics and Nuclear Chemistry question

2010 · Shift 1 · Q12

JEE AdvancedChemistryChemical Kinetics and Nuclear ChemistryNumerical+3 / −1
The concentration of R in the reaction R →\to→ P was measured as a function of time and the following data is obtained
[R] molar 1.0 0.75 0.40 0.10
t (min.) 0.0 0.05 0.12 0.18
The order of reaction is
Numerical answer
View written solutionFree

Correct answer: 0

  1. We test which integrated rate law fits the data.

Given data:

[R]1.000.750.400.10t (min)0.000.050.120.18\begin{array}{c|cccc} {[R]} & 1.00 & 0.75 & 0.40 & 0.10 \\ t\,(\text{min}) & 0.00 & 0.05 & 0.12 & 0.18 \end{array}[R]t(min)​1.000.00​0.750.05​0.400.12​0.100.18​
  1. For a zero-order reaction:
[R]=[R]0−kt[R] = [R]_0 - kt[R]=[R]0​−kt

So, [R][R][R] should decrease linearly with ttt, and

k=[R]0−[R]tk = \frac{[R]_0-[R]}{t}k=t[R]0​−[R]​

Let us calculate kkk from different data points using [R]0=1.0[R]_0=1.0[R]0​=1.0 at t=0t=0t=0.

  • At t=0.05t=0.05t=0.05 min, [R]=0.75[R]=0.75[R]=0.75:
k=1.0−0.750.05=0.250.05=5k = \frac{1.0-0.75}{0.05} = \frac{0.25}{0.05} = 5k=0.051.0−0.75​=0.050.25​=5
  • At t=0.12t=0.12t=0.12 min, [R]=0.40[R]=0.40[R]=0.40:
k=1.0−0.400.12=0.600.12=5k = \frac{1.0-0.40}{0.12} = \frac{0.60}{0.12} = 5k=0.121.0−0.40​=0.120.60​=5
  • At t=0.18t=0.18t=0.18 min, [R]=0.10[R]=0.10[R]=0.10:
k=1.0−0.100.18=0.900.18=5k = \frac{1.0-0.10}{0.18} = \frac{0.90}{0.18} = 5k=0.181.0−0.10​=0.180.90​=5

Since kkk is constant, the reaction follows zero-order kinetics.

  1. Therefore, the order of reaction is:
0\boxed{0}0​
  1. Comparison with stored correct answer:

Stored correct answer = 000

Our derived answer = 000

They match.

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