Question

The unit of rate constant for a zero order reaction.
(A) $mol\quad L^{ -1 }\quad { s }^{ -1 }$
(B) $L\quad mol^{ -1 }\quad s^{ -1 }$
(C) $L^{ 2 }\quad mol^{ -2 }\quad s^{ -1 }$
(D) ${ s }^{ -1 }$

Answer 1

Hint: We know that order of the reaction is the sum of powers of the chemical species that are involved in the rate law. Substitute the given order of the reaction and find out the integrated rate law equation. Then we can find expressions for rate constant and units can also be known.

Complete step by step answer:
> We know that rate law is determined experimentally so the order of reaction can be calculated after experimenting. Let us consider the reaction involving single species and calculate integrated rate law for that reaction.
> Let Reaction be: $\left[ A \right] \quad \rightarrow \quad products$
Let x be the order of the reaction. Then the rate will be proportional to. $\left[ A \right] ^{ x }$. $\left[ A \right]$ is the concentration of the reactant. x is the order of the reaction.
> If we consider some proportionality constant k which is nothing but rate constant of the reaction. Then differential rate law will be:
\[-\dfrac { d\left[ A \right] }{ dt } \quad =\quad k\left[ A \right] ^{ x } \]
> We are given that the reaction is zero-order reaction so x = 0. We substitute this in the rate law equation.
\[\dfrac { d\left[ A \right] }{ dt } \quad =\quad -k\left[ A \right] ^{ 0 } \]
\[\dfrac { d\left[ A \right] }{ dt } \quad =\quad -k\]
\[d\left[ A \right] \quad =\quad -k\quad dt\]
Integrate on both sides we get integrated rate law.
\[\int { d\left[ A \right] \quad =\quad -\int { k } dt }\]
\[\left[ A \right] \quad =\quad -kt\quad +\quad c\]
Where c is integration constant.
\[\left[ A \right] \quad -\quad c\quad =\quad -kt\]
\[k\quad =\quad -\dfrac { \left[ A \right] \quad -\quad c }{ t } \]
\[k\quad =\quad \dfrac { c\quad -\quad \left[ A \right] }{ t }\]
The numerator has units of concentration and the denominator has units of time so fraction will have units of concentration per time.
Units of k = $\dfrac { mol\quad { L }^{ -1 } }{ s } \quad =\quad mol\quad { L }^{ -1 }\quad { s }^{ -1 }$
Therefore, option A is correct.

Note: Integration constant can be obtained by substituting initial concentration and time = 0. We considered the reaction of single chemical species so that integrated rate law can be derived easily. Some zero order reactions depend on the concentration of reactants.