The first order reaction A_{(g)} rightarrow B | vedclass

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Question

(C) For the reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$,let the initial concentration of $A$ be $100 \ M$ at $t = 0$.At $t = 24 \ min$,let the c

Vedclass · Accepted Answer

$157.8$

Vedclass · Answer

(C) For the reaction $A_{(g)} ightarrow B_{(g)} + 2C_{(g)}$,let the initial concentration of $A$ be $100 \ M$ at $t = 0$.At $t = 24 \ min$,let the concentration of $A$ reacted be $x$.Concentration of $A$ remaining $= 100 - x$.Concentration of $B = x$ and concentration of $C = 2x$.Total concentration of products $= x + 2x = 3x$.Given the ratio of products to reactant is $1:3$,so $\frac{3x}{100 - x} = \frac{1}{3}$.$9x = 100 - x \implies 10x = 100 \implies x = 10$.Concentration of $A$ remaining $[A]_t = 100 - 10 = 90$.Rate constant $k = \frac{2.303}{t} \log(\frac{[A]_0}{[A]_t}) = \frac{2.303}{24} \log(\frac{100}{90}) = \frac{2.303}{24} \log(1.11)$.Using $\log 1.11 = 0.046$,$k = \frac{2.303 	imes 0.046}{24} \approx 0.004415 \ min^{-1}$.Half-life $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.004415} \approx 157.19 \ min$,which is approximately $157.8 \ min$.

The first order reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$ occurs at $25^{\circ} C$. After $24 \ min$,the ratio of the concentration of products to the concentration of the reactant is $1:3$. What is the half-life of the reaction (in $min$)? $(\log 1.11 = 0.046)$

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Assertion : For a first order reaction,$t_{1/2}$ is independent of initial concentration.
Reason : For a first order reaction,rate constant $k \propto [R]$.

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