
What are the units of rate constant for the first order reaction?
Answer
579.3k+ views
Hint: We know that rate of a reaction is explained as change in concentration of any of the reactants or product per unit time. For example:
${\text{aA + bB}} \to {\text{cC + dD}}$
Rate of reaction is equal to decrease in concentration of either A or B or increases in concentration of C or D per unit time. In the above reaction a, b, c, and d are stoichiometric coefficients. Thus rate for the above general reaction is equal to rate of removal of A or B per mole and also equal to rate appearance of C or D per mole. It can be shown is as follows:
$
{\text{rate}} = - \dfrac{{\text{1}}}{{\text{a}}}\dfrac{{{\text{d}}\left[ {\text{A}} \right]}}{{{\text{dt}}}} = - \dfrac{{\text{1}}}{{\text{b}}}\dfrac{{{\text{d}}\left[ {\text{B}} \right]}}{{{\text{dt}}}} \\
\,\,\,\,\,\,\,\,\, = + \dfrac{{\text{1}}}{{\text{c}}}\dfrac{{{\text{d}}\left[ {\text{C}} \right]}}{{{\text{dt}}}}{\text{ = + }}\dfrac{{\text{1}}}{{\text{d}}}\dfrac{{{\text{d}}\left[ {\text{D}} \right]}}{{{\text{dt}}}} \\
$
Here [ ] represents concentration in mole per liter and ‘d’ represents an infinitesimally small change in concentration. Negative sign shows that concentration of reactant A and B are decreasing whereas positive sign shows concentration of product C and D are increasing.
Complete step by step answer:
As we know a first order reaction is represented as
${\text{A}} \to \,\,\,\,\,\,\,{\text{Product}}$
Initial concentration: $a$ $0$
Concentration after time: $\left( {a - x} \right)$ $x$
Now we put differential rate law:
$ - \dfrac{{d\left( {a - x} \right)}}{{dt}} = + \dfrac{{dx}}{{dt}} = {k_1}\left( {a - x} \right)$
On integrating above equation we get
Integration rate law:
${k_1} = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)$
Here ${k_1}$ is the rate constant of a first order reaction unit that is per time or ${{\text{s}}^{ - 1}}$ since rest of the expression consists of numerical values and log does not have any units.
Thus, the unit of first order reaction is ${{\text{s}}^{ - 1}}$ .
Note:
Rate of the reaction is proportional to the product of concentration of reactants, each raised to some power. i.e.
$
{\text{rate}} \propto {\left[ {\text{A}} \right]^{\text{m}}}{\left[ {\text{B}} \right]^{\text{n}}} \\
{\text{rate = k}}{\left[ {\text{A}} \right]^{\text{m}}}{\left[ {\text{B}} \right]^{\text{n}}} \\
$
Here k is rate constant. At all concentrations, rate constant k is equal to rate of a reaction.
${\text{aA + bB}} \to {\text{cC + dD}}$
Rate of reaction is equal to decrease in concentration of either A or B or increases in concentration of C or D per unit time. In the above reaction a, b, c, and d are stoichiometric coefficients. Thus rate for the above general reaction is equal to rate of removal of A or B per mole and also equal to rate appearance of C or D per mole. It can be shown is as follows:
$
{\text{rate}} = - \dfrac{{\text{1}}}{{\text{a}}}\dfrac{{{\text{d}}\left[ {\text{A}} \right]}}{{{\text{dt}}}} = - \dfrac{{\text{1}}}{{\text{b}}}\dfrac{{{\text{d}}\left[ {\text{B}} \right]}}{{{\text{dt}}}} \\
\,\,\,\,\,\,\,\,\, = + \dfrac{{\text{1}}}{{\text{c}}}\dfrac{{{\text{d}}\left[ {\text{C}} \right]}}{{{\text{dt}}}}{\text{ = + }}\dfrac{{\text{1}}}{{\text{d}}}\dfrac{{{\text{d}}\left[ {\text{D}} \right]}}{{{\text{dt}}}} \\
$
Here [ ] represents concentration in mole per liter and ‘d’ represents an infinitesimally small change in concentration. Negative sign shows that concentration of reactant A and B are decreasing whereas positive sign shows concentration of product C and D are increasing.
Complete step by step answer:
As we know a first order reaction is represented as
${\text{A}} \to \,\,\,\,\,\,\,{\text{Product}}$
Initial concentration: $a$ $0$
Concentration after time: $\left( {a - x} \right)$ $x$
Now we put differential rate law:
$ - \dfrac{{d\left( {a - x} \right)}}{{dt}} = + \dfrac{{dx}}{{dt}} = {k_1}\left( {a - x} \right)$
On integrating above equation we get
Integration rate law:
${k_1} = \dfrac{{2.303}}{t}\log \left( {\dfrac{a}{{a - x}}} \right)$
Here ${k_1}$ is the rate constant of a first order reaction unit that is per time or ${{\text{s}}^{ - 1}}$ since rest of the expression consists of numerical values and log does not have any units.
Thus, the unit of first order reaction is ${{\text{s}}^{ - 1}}$ .
Note:
Rate of the reaction is proportional to the product of concentration of reactants, each raised to some power. i.e.
$
{\text{rate}} \propto {\left[ {\text{A}} \right]^{\text{m}}}{\left[ {\text{B}} \right]^{\text{n}}} \\
{\text{rate = k}}{\left[ {\text{A}} \right]^{\text{m}}}{\left[ {\text{B}} \right]^{\text{n}}} \\
$
Here k is rate constant. At all concentrations, rate constant k is equal to rate of a reaction.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

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

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

State the principle of an ac generator and explain class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Sketch the electric field lines in case of an electric class 12 physics CBSE

Give 10 examples of unisexual and bisexual flowers

