Question

For the zero order reaction ${\rm{A}} \to 2{\rm{B}}$, the rate constant is $2 \times {10^{ - 6}}\;{\rm{M}}{\rm{.mi}}{{\rm{n}}^{ - 1}}$. The reaction is started with ${\rm{10}}\;{\rm{M}}$ A.
(i). What will be the concentration of A after $2$ days (ii) What is the initial half-life of the reaction.
(iii) In what time, the reaction will complete?

Answer 1

We know that the half-life of the zero order reaction will decrease with the initial concentration of that reaction decreasing.

Complete step by step solution
Given the rate constant is $2 \times {10^{ - 6}}\;{\rm{M}}{\rm{.mi}}{{\rm{n}}^{ - 1}}$.
The concentration is ${\rm{10}}\;{\rm{M}}$.
The zero order reaction ${\rm{A}} \to 2{\rm{B}}$.
(i). The formula for concentration of zero order reaction is shown below
${{\rm{K}}_{\rm{t}}} = {{\rm{a}}_{\rm{0}}} - {{\rm{a}}_{\rm{t}}}$
Where, ${{\rm{K}}_{\rm{t}}}$ is the rate constant with respect to time t, ${{\rm{a}}_{\rm{0}}}$ is the initial concentration, and ${{\rm{a}}_{\rm{t}}}$ is the concentration at A.
In the question ${\rm{2}}\;{\rm{days}}$ are given so, conversion of days to min is done as follows.
${\rm{t}} = {\rm{2days}}\\\ = 42 \times 24 \times 60\\\ = 2880\;{\rm{min}}$
Substitute the value in the above equation as shown below.
$\Rightarrow 2 \times {10^{ - 6}}\;{\rm{M}}{\rm{.mi}}{{\rm{n}}^{ - 1}} \times 2880\;\min = \Rightarrow 10\;{\rm{M}} - {{\rm{a}}_{\rm{t}}}\\\ \Rightarrow {{\rm{a}}_{\rm{t}}} = 10\;{\rm{M}} - 0.00576\;{\rm{M}}\\\ = 9.99424\;{\rm{M}}$
Hence, the concentration of A after ${\rm{2}}\;{\rm{days}}$ is $9.99424\;{\rm{M}}$.

(ii). For zero order reaction, the initial half-life equation is shown below.
${{\rm{a}}_{\rm{t}}} = \dfrac{{{{\rm{a}}_{\rm{0}}}}}{{\rm{2}}}\\\ {\rm{i}}{\rm{.e}}{\rm{.,}}{{\rm{t}}_{\frac{{\rm{1}}}{{\rm{2}}}}} = \dfrac{{{{\rm{a}}_{\rm{0}}}}}{{{\rm{2K}}}}$
Where, ${{\rm{a}}_{\rm{0}}}$ is the initial concentration, ${{\rm{t}}_{\frac{{\rm{1}}}{{\rm{2}}}}}$ is the half-life for zero order reaction, ${\rm{K}}$ is the rate constant.
Substitute the respective values in the above equation.
$\Rightarrow {{\rm{t}}_{\frac{{\rm{1}}}{{\rm{2}}}}} = \dfrac{{{\rm{10}}}}{{{\rm{2}} \times {\rm{1}} \times {\rm{1}}{{\rm{0}}^{ - 6}}}}\\\ = 2.5 \times {10^6}\;{\rm{min}}$
Hence, the half-life for zero order reaction is $2.5 \times {10^6}\;{\rm{min}}$.

(iii). The reaction will be completed in which time can be calculated by using the formula as shown below.
${\rm{t}} = \dfrac{{{{\rm{a}}_{\rm{0}}}}}{{\rm{K}}}$
Where t is the time in min, K is the rate constant and ${{\rm{a}}_{\rm{0}}}$ is the initial concentration.
Substitute the respective values in the above equation. We get,
${\rm{t}} = \dfrac{{{\rm{10}}}}{{2 \times {{10}^{ - 6}}}}\\\ = 5 \times {10^6}\;{\rm{min}}$
Hence, the reaction will complete in $5 \times {10^6}\;{\rm{min}}$.

Note:
The main application of rate of reaction generally represents or shows the rate of production which was used in our daily life.