Question

Consider the reaction, \[{\text{C + D}} \to {\text{Product}}\] . The rate of reaction increases by a factor of \[4\] when the concentration of \[{\text{C}}\] is doubled. The rate of reaction is tripled when concentration of \[{\text{D}}\] is tripled. What is the order of reaction? Write rate law.

Answer 1

Hint: To answer this question we should know the rate of reaction, order of reaction, rate law, factors affecting the rate of reaction. Rate of reaction comes under chemical kinetics which is a branch of physical chemistry and deals with the rate of chemical reaction and explains in which direction the reaction is taking place either forward or backward.

Complete step by step answer:
We know that the rate of reaction is also known as reaction rate. It is defined as the speed at which the reaction occurs or the rate at which the reactants are converted to products. Rate can be determined by dividing the concentration with time.
Now we can see the factors affecting the rate of reaction as:
Rate is proportional to concentration.
Rate is proportional to temperature.
Presence of a catalyst can speed up or slow down the reaction rate.
Rate is proportional to surface area (i.e. small size)
Unit of rate of reaction is \[{\text{mol/L/s}}\]
Rate law: also known as rate equation. It is a relation between rate of reaction and the concentration of reactants taking part in the reaction. It is expressed as;
\[{\text{Rate = k[A}}{{\text{]}}^{\text{x}}}{{\text{[B]}}^{\text{y}}}{{\text{[C]}}^{\text{z}}}\]
Where k is rate constant or proportionality constant, \[{\text{[A],[B],[C]}}\] represents concentration of reactants and exponents x, y, z represents the order of reaction which is calculated by adding up all the exponents.
In this question, we have following equation,
\[{\text{C + D}} \to {\text{Product}}\]
So,
\[{\text{Rat}}{{\text{e}}_{\text{1}}}{\text{ = K[C}}{{\text{]}}^{\text{x}}}{{\text{[D]}}^{\text{y}}}\] ………. (1)
Let the order of reaction be x and y for reactants C and D respectively. Overall order of reaction is \[{\text{x + y}}\]now, as per the question rate increases by factor 4 so we have
 \[{\text{4}}{{\text{R}}_{\text{1}}}{\text{ = k[2C}}{{\text{]}}^{x{\text{ }}}}{{\text{[D]}}^y}\] ……. (2)
Rate of reaction is tripled if concentration of D is tripled, then;
\[{\text{3}}{{\text{R}}_{\text{1}}}{\text{ = k[C}}{{\text{]}}^{x{\text{ }}}}{{\text{[3D]}}^y}\] …… (3)
\[\therefore {\text{ 2 - 1}}\] (Subtract equation 1 from 2) we get,
\[\dfrac{{{\text{4}}{{\text{R}}_{\text{1}}}}}{{{{\text{R}}_{\text{1}}}}}{\text{ = }}{\dfrac{{{\text{k[2C}}{{\text{]}}^{\text{x}}}{\text{[D]}}}}{{{\text{k[C}}{{\text{]}}^{\text{x}}}{{{\text{[D]}}}^{\text{y}}}}}^{\text{y}}}\]
\[ \Rightarrow \dfrac{{{\text{4}}{{\text{R}}_{\text{1}}}}}{{{{\text{R}}_{\text{1}}}}}{\text{ = [2}}{{\text{]}}^{\text{x}}}\]
\[{{\text{2}}^2}{\text{ = }}{{\text{2}}^{\text{x}}}\]
On simplification we get,
\[{\text{x = 2}}\]
Now,
\[3{\text{ - 1}}\] (Subtract equation 1 from 3) we get,
\[\dfrac{{{\text{3[}}{{\text{R}}_{\text{1}}}{\text{]}}}}{{{{\text{R}}_{\text{1}}}}}{\text{ = }}\dfrac{{{\text{k[C}}{{\text{]}}^{\text{x}}}{{{\text{[3D]}}}^{\text{y}}}}}{{{\text{k[C}}{{\text{]}}^{{\text{x }}}}{{{\text{[D]}}}^{\text{y}}}}}\]
\[{{\text{3}}^1}{\text{ = }}{{\text{3}}^{\text{y}}}\]
On simplification we get,
\[ \Rightarrow {\text{y = 1}}\]
So rate law is
\[{{\text{R}}_{{\text{1 }}}}{\text{ = k[C}}{{\text{]}}^2}{\text{[D]}}\]
Overall order of reaction is 3.

Note: As we know that the rate law can be determined experimentally only. The proportionality constant k does not depend on the concentration but is affected by change in temperature and surface area. Order of reactions can be zero, first, second and third. It is to be noted that the rate of reaction is always positive and if it is negative then it shows the reactants concentration is decreasing.