Question

What will be the half – life, ${t^{\dfrac{1}{2}}}$ if the 99% of the first order reaction is completed in 330 minutes?
A) 60 min
B) 50 min
C) 70 min
D) 40 min

Answer 1

Hint: 99% of the initial concentration is converted into product. Calculate the amount of concentration left after the conversion of reactants into products. Find the rate constant using the rate constant formula of first order reaction. Then, use half – life formula for first order reaction.

Formula used: The rate constant of the first order reaction is given by –
$k = \dfrac{{2.303}}{t}\log \dfrac{{\left[ {{R_1}} \right]}}{{\left[ {{R_2}} \right]}}$
where, $t$ is the life taken by the reaction to complete
$\left[ {{R_1}} \right]$ is the initial concentration of the reaction
$\left[ {{R_2}} \right]$ is the left concentration of the reaction after its conversion from reactants into products.
The half – life for first order reaction is given by –
${t^{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$

Complete step by step solution:
From the question, it is given that 99% of the initial concentration of the reaction is converted into product.
The life taken by the reaction to complete, $t = 330\min $
Given that the reaction is the first – order reaction.
Let the initial concentration of the reaction be $a$ and let the reacted amount be $x$.
Therefore, the reacted amount will be –
$
   \Rightarrow x = \dfrac{{a \times 99}}{{100}} \\
   \Rightarrow x = 0.99a \\
 $
Then, the left amount from the reaction is $ \Rightarrow a - 0.99a = 0.01a$
Now, we know that, for first order reaction the rate constant is given by –
$k = \dfrac{{2.303}}{t}\log \dfrac{{\left[ {{R_1}} \right]}}{{\left[ {{R_2}} \right]}} \cdots \left( 1 \right)$
where, $t$ is the life taken by the reaction to complete
$\left[ {{R_1}} \right]$ is the initial concentration of the reaction
$\left[ {{R_2}} \right]$ is the left concentration of the reaction after its conversion from reactants into products.
$\therefore k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}$
Putting the values in the above equation –
$
   \Rightarrow k = \dfrac{{2.303}}{{330}}\log \dfrac{a}{{0.01a}} \\
   \Rightarrow k = \dfrac{{2.303}}{{330}}\log \dfrac{{100}}{1} \\
   \Rightarrow k = \dfrac{{2.303}}{{330}} \times 2 \\
  \therefore k = 0.0139 \\
 $
We know that, the formula of half – life for first order reaction is –
${t^{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
Putting the value of rate constant –
$
   \Rightarrow {t^{\dfrac{1}{2}}} = \dfrac{{0.693}}{{0.0139}} \\
   \Rightarrow {t^{\dfrac{1}{2}}} = 49.85 \\
   \Rightarrow {t^{\dfrac{1}{2}}} \approx 50\min \\
 $
Hence, the half – life of the reaction will be approx. 50 minutes.

Hence the correct answer is option ‘B’.

Note: Half life is the time required by the quantity to reduce to its half of the initial value. Half-life is constant over the lifetime of an exponentially decaying quantity, and it is a characteristic unit for the exponential decay equation.