Answer
Verified
35.7k+ views
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.
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.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Three point particles of mass 1 kg 15 kg and 25 kg class 11 physics JEE_Main
Which of the following explanation is best for not class 12 chemistry JEE_Main
Other Pages
Two identical charged spheres suspended from a common class 12 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main