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Question: How do coefficients affect the rate law?...

How do coefficients affect the rate law?

Explanation

Solution

You must have the knowledge of rate law and order of reaction to determine the answer. In chemical kinetics the reaction rate constant or the reaction rate coefficient denoted by k expresses the rate and the direction of the chemical reaction.

Complete step by step answer:
The rate law is also known as the rate equation is a mathematical expression which describes the relationship between the rate of a chemical reaction and the concentration of the reactants.
The general rate law is given as shown below.
rate=k[A]m[B]n[C]prate = k{[A]^m}{[B]^n}{[C]^p}
Where,
[A], [B], [C] is the molar concentration of the reactants.
K is the rate constant
The exponents m, n and p are positive integers.
To determine how the coefficients affect the rate law, let’s take the example.
N2O42NO2{N_2}{O_4} \rightleftharpoons 2N{O_2}
Each coefficient present worked into the rate of appearance/ disappearance
1υd[A]dt=1υd[B]dt=r(t)=k[A]order- \dfrac{1}{\upsilon }\dfrac{{d[A]}}{{dt}} = \dfrac{1}{\upsilon }\dfrac{{d[B]}}{{dt}} = r(t) = k{[A]^{order}}
Where,
υ\upsilon is the stoichiometric coefficient
A is the reactant
B is the product
In this reaction, N2O4{N_2}{O_4} is the decomposing reactant, therefore it is a first order reaction.
So, we get
d[N2O4]dt=12d[NO2]dt=r(t)=k[N2O4]1- \dfrac{{d[{N_2}{O_4}]}}{{dt}} = \dfrac{1}{2}\dfrac{{d[N{O_2}]}}{{dt}} = r(t) = k{[{N_2}{O_4}]^1}
Hence, the stoichiometric coefficient does not affect how the rate law should be written. But the coefficient affects the value of rate constant K.
The value of the coefficient k changes with the conditions that affect reaction rate, like temperature, pressure, surface area, etc. A smaller rate constant value shows that the reaction is slow, while a larger rate constant indicates a faster reaction.

Note:
The reaction order is not related to the stoichiometric coefficients. The reaction order only determines the number of molecules per mole participating in the reaction, not how many moles of each molecule are there.