Solveeit Logo

Question

Question: The rate of first order reaction is \(0.04\text{ }molL/s\) at 10 minutes and \(0.03\text{ }molL/s\) ...

The rate of first order reaction is 0.04 molL/s0.04\text{ }molL/s at 10 minutes and 0.03 molL/s0.03\text{ }molL/s at 20 minutes after initiation. Find the half-life of the reaction.
A. 2.406 min
B. 24.06 min
C. 240.6 min
D. 0.204 min

Explanation

Solution

Try to recall that the half-life of a reaction is time duration during which the concentration of a reactant is reduced to half of its initial concentration. Also, the half period of a first order reaction is independent of initial concentration. Now by using this and the formula given below you can easily find the half-life of reaction.

Complete step by step answer:
Recall the equation k=2.303tlog10aAk=\dfrac{2.303}{t}{{\log }_{10}}\dfrac{a}{A}, where kk is the rate constant for first order reaction, aa is the initial concentration and AA is the final concentration.
t1/2=0.693k{{t}_{1/2}}=\dfrac{0.693}{k}, where kk is the rate constant and t1/2{{t}_{1/2}} is the half-life for the first order reaction.
- The rate law expression for a first order reaction rate is given, Rate=k[A]Rate=k[A] where [A][A] is the concentration of reactant at time tt.

Calculations:
Let r1{{r}_{1}} the rate of reaction after 10 min. and r2{{r}_{2}} be the rate of reaction after 20 min.
So, r1=k[A]{{r}_{1}}=k\left[ A \right] and r2=k[B]{{r}_{2}}=k\left[ B \right].
Given that,

& {{r}_{1}}=0.04\text{ }molL/s \\\ & {{r}_{2}}=0.03\text{ }molL/s \\\ \end{aligned}$$ $${{r}_{1}}=k\left[ A \right]$$ or, Eq.1: $0.04=k\left[ A \right]$ Similarly, ${{r}_{2}}=k\left[ B \right]$ or, Eq.2: $0.03=k\left[ B \right]$ Dividing eq. 1 by eq. 2, we get $$\dfrac{\left[ A \right]}{\left[ B \right]}=\dfrac{4}{3}$$ Let $[A]$ be the initial concentration of reactant and $[B]$ be the final concentration after a time gap of $t=10\min $ So, using above formula $$\begin{aligned} & k=\dfrac{2.303}{t}{{\log }_{10}}\dfrac{\left[ A \right]}{\left[ B \right]} \\\ & k=\dfrac{2.303}{10}{{\log }_{10}}\dfrac{4}{3} \\\ & \Rightarrow k=0.0285{{s}^{-1}} \\\ \end{aligned}$$ We know that, $$\begin{aligned} & {{t}_{1/2}}=\dfrac{0.693}{k} \\\ & \Rightarrow {{t}_{1/2}}=\dfrac{0.693}{0.0285} \\\ & or,\text{ }{{t}_{1/2}}=24.06\min \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** It should be remembered to you that radioactive disintegration reactions are first order reactions. In these reactions, the whole of substance never disintegrates. The formula for the half-life of any species changes as per the order of the reaction.