Question

A reaction of first-order got completed $90\%$ in $90$ minutes. Hence it will complete $50\%$ in approximately:
a) $50\min$
b) $54\min$
c) $27\min$
d) $62\min$

Answer 1

With the help of rate law expression we can calculate the time required for the completion of a certain percentage of the reaction . The given reaction is first order which means the rate of reaction depends only on the concentration of one reactant.

Complete step by step answer:
Order of a reaction: It is termed as the sum of power of concentration of reactants in rate law expression.
Rate law expression is given as follows:
Rate law expression is given as follows:
Rate= $K{\left[ A \right]^x}{\left[ B \right]^y}$
${\left[ A \right]^x} =$ concentration of reactant A
${\left[ B \right]^y} =$ concentration of reactant B
There are different types of reactions namely: zero order reaction, first order reaction, second order reaction. All of these reactions depend upon the concentration of the reactants.
We will learn about each type of reaction in detail:
Zero order reaction:
It is the type of reaction in which the rate of reaction does not depend on the concentration of the reactants.
First order reaction
It is the type of reaction in which rate of reaction depends upon the concentration of one reactant only.
Second order reaction:
It is the type of reaction in which rate of reaction depends upon the concentration of two reactants.
Rate law equation:
The reaction for the first order will be given as: $A + 2B \to C$
So the rate law for the first order reaction will be given as: rate$= k\left[ A \right]$.
The rate equation for the first order reaction is:
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
Where k= rate constant
T= time
${\left[ A \right]_0} =$ Concentration at initial time
${\left[ A \right]_t} =$ Concentration at time t.
Half-life for the first order reaction: it is defined as the time required by half of the initial concentration of reactant to be completed by the reaction.
So it is given by the equation:
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
$t\dfrac{1}{2} =$ half-life of the first order reaction
K $=$ rate constant
Now we will use all the above equations of rate law and half-life to solve the sum.

Given data:
Time: $90$mins.
Work completed in time t= $90\%$
So now we will use the rate law equation, $k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
Substituting the value in rate law equation we get,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
${\left[ A \right]_0} =$ Total concentration
${\left[ A \right]_t}$ = amount of reaction that got completed.
So it will be given as,
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_0} - {{\left[ A \right]}_t}}}$

substituting the value we get,
$k = \dfrac{{2.303}}{{90}}\log \dfrac{{100}}{{100 - 90}}$
$\Rightarrow$ $k = \dfrac{{2.303}}{{90}}\log \dfrac{{100}}{{10}}$
$\Rightarrow$ $k = \dfrac{{2.303}}{{90}}\log 10$
$\Rightarrow$ $k = \dfrac{{2.303}}{{90}} \times 1$
$k = 0.0255$min-1
We can calculate in two different ways given below:
By using the rate law equation and value of k from the above solution
Now will calculate the half life using the equation: ${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}$
Substituting the values, we get:
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{0.0255}}$
${t_{\dfrac{1}{2}}} = 27.18$

Rate law equation when reaction is completed in time ‘t1’ we get:
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_0} - {{\left[ A \right]}_t}}}$
Substituting the values of k $= 0.0255$ and ${\left[ A \right]_t} = 50\%$ we get,
${t_1} = \dfrac{{2.303}}{{0.0255}}\log \dfrac{{100}}{{100 - 50}}$
$\Rightarrow$ ${t_1} = \dfrac{{2.303}}{{0.0255}}\log \dfrac{{100}}{{50}}$
$\Rightarrow$ ${t_1} = 90.313\log 2$
$\Rightarrow$ ${t_1} = 90.313 \times 0.301$
${t_1} = 27.18$mins

So the time taken to complete $50\%$ reaction is $27.18$ min$\approx 27$ mins . So the answer for this option c) $27$ mins.

Note: For the reversible first order reaction sometimes the amount of product and reactants will be the same, then it will be in equilibrium. So If the reaction will be in equilibrium their rate constants will also be in equilibrium.