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Chemical Kinetics and Nuclear Chemistry question

2022 · 26 Jul · Shift 1 · Q22
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Chemical Kinetics and Nuclear Chemistry question

2022 · 26 Jul · Shift 1 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1
For a reaction A→2 B+C\mathrm{A} \rightarrow 2 \mathrm{~B}+\mathrm{C}A→2 B+C the half lives are 100 s100 \mathrm{~s}100 s and 50 s50 \mathrm{~s}50 s when the concentration of reactant A\mathrm{A}A is 0.50.50.5 and 1.0 mol L−11.0 \mathrm{~mol} \mathrm{~L}^{-1}1.0 mol L−1 respectively. The order of the reaction is ‾\underline{\hspace{2cm}}​ . (Nearest Integer)
Numerical answer
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Correct answer: 2

  1. For an nthn^{\text{th}}nth order reaction (n≠1)(n\neq 1)(n=1), the half-life is given by
t1/2∝1[A]0 n−1t_{1/2} \propto \frac{1}{[A]_0^{\,n-1}}t1/2​∝[A]0n−1​1​

More explicitly,

t1/2=2n−1−1(n−1)k[A]0n−1t_{1/2}=\frac{2^{n-1}-1}{(n-1)k[A]_0^{n-1}}t1/2​=(n−1)k[A]0n−1​2n−1−1​

So, for the same reaction,

t1/2,1t1/2,2=([A]0,2[A]0,1)n−1\frac{t_{1/2,1}}{t_{1/2,2}}=\left(\frac{[A]_{0,2}}{[A]_{0,1}}\right)^{n-1}t1/2,2​t1/2,1​​=([A]0,1​[A]0,2​​)n−1
  1. Given:
t1/2,1=100 s at [A]0,1=0.5 mol L−1t_{1/2,1}=100\ \text{s at } [A]_{0,1}=0.5\ \text{mol L}^{-1}t1/2,1​=100 s at [A]0,1​=0.5 mol L−1 t1/2,2=50 s at [A]0,2=1.0 mol L−1t_{1/2,2}=50\ \text{s at } [A]_{0,2}=1.0\ \text{mol L}^{-1}t1/2,2​=50 s at [A]0,2​=1.0 mol L−1

Thus,

10050=(1.00.5)n−1\frac{100}{50}=\left(\frac{1.0}{0.5}\right)^{n-1}50100​=(0.51.0​)n−1 2=2n−12=2^{n-1}2=2n−1
  1. Comparing powers of 222:
n−1=1n-1=1n−1=1 n=2n=2n=2
  1. Therefore, the order of the reaction is
2\boxed{2}2​
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