Question

What is the two third of a first order reaction having $\text{K}=5.48\times {{10}^{-14}}\text{ }{{\text{s}}^{-1}}$?

Answer 1

A chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance. Examples of first order reactions are
$\text{S}{{\text{O}}_{2}}\text{C}{{\text{l}}_{2}}\to \text{C}{{\text{l}}_{2}}\text{+S}{{\text{O}}_{2}}$
$2{{\text{N}}_{2}}{{\text{O}}_{5}}\to {{\text{O}}_{2}}+4\text{N}{{\text{O}}_{2}}$
Formula used - ${{\text{t}}_{\text{n}}}=\dfrac{2.303}{\text{k}}\log \dfrac{\left[ {{\text{A}}_{0}} \right]}{\left[ \text{A} \right]}$

Complete step by step answer:
According to this question, the two third of first order reaction having $\text{K}=5.48\times {{10}^{-14}}{{\text{s}}^{-1}}$-14 is $2\times {{10}^{13}}\text{s}$.
Step by step answer–According to the question a first order reaction is a chemical reaction in which reaction rate is dependent on the concentration of only one reactant. In this question we have to find two third of the first order reaction.
In this question value of k is given i.e. $\text{K}=5.48\times {{10}^{-14}}\text{ }{{\text{s}}^{-1}}$
$\Rightarrow$ ${{\text{t}}_{\text{n}}}=\dfrac{2.303}{\text{k}}\log \dfrac{\left[ {{\text{A}}_{0}} \right]}{\left[ \text{A} \right]}$
[A] is the current concentration of the first-order reactant
${{\left[ \text{A} \right]}_{0}}$is the initial concentration of the first-order reactant
t is the time elapsed since the reaction began
k is the rate constant of the first-order reaction
Initial value of the first order reactant is $\left[ {{\text{A}}_{0}} \right]=\text{a}$
The current concentration of the first order reactant is
$\Rightarrow$ $\left[ \text{A} \right]=\text{a}-\dfrac{2}{3}\text{a}$
$\Rightarrow$ $\left[ \text{A} \right]=\dfrac{\text{a}}{3}$
$\Rightarrow$ ${{\text{t}}_{\text{n}}}$ is the time taken for two third reaction is
$\Rightarrow$ ${{\text{t}}_{\text{n}}}=\text{t}_{2/3}$
K is the rate constant of first order reaction
$\Rightarrow$ $\text{K}=5.48\times {{10}^{-14}}\text{ }{{\text{s}}^{-1}}$
Now will we substitute the values in the formula i.e.
$\Rightarrow$ ${{\text{t}}_{\text{n}}}=\dfrac{2.303}{\text{k}}\log \dfrac{\left[ {{\text{A}}_{0}} \right]}{\left[ \text{A} \right]}$
$\text{t}_{{2}/{3}}=\dfrac{2.303}{5.48\times {{10}^{-14}}}\log 3$
Value of log 3 is
Log 3 = 0.477
$\Rightarrow$ $\text{t}_{{2}/{3}}=\dfrac{2.303}{5.48\times {{10}^{-14}}}\times 0.477$
$\Rightarrow$ $\text{t}_{{2}/{3}}=\dfrac{2.303\times 0.477\times {{10}^{14}}}{5.48}$
$\Rightarrow$ $\text{t}_{{2}/{3}}=\dfrac{1.098\times {{10}^{14}}}{5.48}$
$\Rightarrow$ $\text{t}_{{2}/{3}}=0.2\times {{10}^{14}}$

$t{}_{{2}/{3}}=2\times {{10}^{13}}s$

Note:
A first-order reaction can be defined as a chemical reaction in which the reaction rate is linearly dependent on the concentration of only one reactant. In other words, a first-order reaction is a chemical reaction in which the rate varies based on the changes in the concentration of only one of the reactants. Thus, the order of these reactions is equal to 1.