
A first- order is $20\% $ complete in 10 minutes. Calculate the rate constant of the reaction
(II) Time taken for the reaction to go to $75\% $ completion.
Answer
528.8k+ views
Hint:To answer this question, you must recall the rate law equation for a first order reaction. Rate constant of a first order reaction does not depend on the concentration of the reactant. We shall substitute the values in the given equation.
Formula used: $kt = \ln \left( {\dfrac{{{a_i}}}{{{a_t}}}} \right)$
Where, $k$ is the rate constant of the first order reaction under consideration
${a_i}$ is the initial concentration of the reactants in the reaction mixture
And ${a_t}$ is the concentration of reactants in the reaction mixture at time $t$
Complete step by step answer:
It is given in the question that the time taken for completion of $20\% $of the reaction $ = 20{\text{ minutes}}$
Let $x$ be the number of moles of reactant used up in the reaction. Assuming the initial concentration of the reactants in the reaction mixture as $100$, we get
$x = 20$
Using the rate law equation, $kt = \ln \left( {\dfrac{{{a_i}}}{{{a_t}}}} \right) = \ln \left( {\dfrac{{{a_i}}}{{{a_i} - x}}} \right)$
Substituting the values, we get,
$k = \dfrac{{2.303}}{{10}}\log \left( {\dfrac{{100}}{{100 - 20}}} \right)$
$ \Rightarrow k = 2.2 \times {10^{ - 2}}{\text{ mi}}{{\text{n}}^{ - 1}}$
(II) Now, let the time taken for $75\% $completion of the reaction be $T$.
Here, $75\% $ of reactant is consumed in the reaction, so $x = 75$
Substituting the values of $x$ and rate constant into the rate law equation, we get,
$T = \dfrac{{2.303}}{{2.2 \times {{10}^{ - 2}}{\text{ }}}}\log \left( {\dfrac{{100}}{{100 - 75}}} \right)$
$\therefore T = 63{\text{ min}}$
Hence, the time taken for $75\% $completion of the reaction is $63{\text{ minutes}}$.
Note:
For a first order reaction, half-life is independent of the initial concentration of the reactant.
The concentration of reactant decreases exponentially with time in a first order equation.
If we are given the value of ${C_0}$ and ${C_t}$ at different time instants, the value of $k$ can be calculated for different time instants by using the first order law. If the reaction for which the data is given is a first order reaction, then all values of $k$ will be approximately equal to each other.
Formula used: $kt = \ln \left( {\dfrac{{{a_i}}}{{{a_t}}}} \right)$
Where, $k$ is the rate constant of the first order reaction under consideration
${a_i}$ is the initial concentration of the reactants in the reaction mixture
And ${a_t}$ is the concentration of reactants in the reaction mixture at time $t$
Complete step by step answer:
It is given in the question that the time taken for completion of $20\% $of the reaction $ = 20{\text{ minutes}}$
Let $x$ be the number of moles of reactant used up in the reaction. Assuming the initial concentration of the reactants in the reaction mixture as $100$, we get
$x = 20$
Using the rate law equation, $kt = \ln \left( {\dfrac{{{a_i}}}{{{a_t}}}} \right) = \ln \left( {\dfrac{{{a_i}}}{{{a_i} - x}}} \right)$
Substituting the values, we get,
$k = \dfrac{{2.303}}{{10}}\log \left( {\dfrac{{100}}{{100 - 20}}} \right)$
$ \Rightarrow k = 2.2 \times {10^{ - 2}}{\text{ mi}}{{\text{n}}^{ - 1}}$
(II) Now, let the time taken for $75\% $completion of the reaction be $T$.
Here, $75\% $ of reactant is consumed in the reaction, so $x = 75$
Substituting the values of $x$ and rate constant into the rate law equation, we get,
$T = \dfrac{{2.303}}{{2.2 \times {{10}^{ - 2}}{\text{ }}}}\log \left( {\dfrac{{100}}{{100 - 75}}} \right)$
$\therefore T = 63{\text{ min}}$
Hence, the time taken for $75\% $completion of the reaction is $63{\text{ minutes}}$.
Note:
For a first order reaction, half-life is independent of the initial concentration of the reactant.
The concentration of reactant decreases exponentially with time in a first order equation.
If we are given the value of ${C_0}$ and ${C_t}$ at different time instants, the value of $k$ can be calculated for different time instants by using the first order law. If the reaction for which the data is given is a first order reaction, then all values of $k$ will be approximately equal to each other.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Why cannot DNA pass through cell membranes class 12 biology CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

