Question

For the reaction: ${\text{2A + B}} \to {{\text{A}}_{\text{2}}}{\text{B}}$ the ${\text{rate = }}k{\text{[A][B}}{{\text{]}}^{\text{2}}}$with\[k = 2.0 \times {10^{ - 6}}{\text{mo}}{{\text{l}}^{{\text{ - 2}}}}{{\text{L}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}}\]. Calculate the initial rate of the reaction when ${\text{[A] = 0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}$ , ${\text{[B] = 0}}{\text{.2 mol }}{{\text{L}}^{{\text{ - 1}}}}$. Calculate the rate of reaction after ${\text{[A]}}$ is reduced to ${\text{0}}{\text{.06 mol }}{{\text{L}}^{{\text{ - 1}}}}$.

Answer 1

Hint: The rate of reaction is the change in concentration of reactant or product in unit time. Change in concentration of reactant affects the rate of reaction. Reaction rate has units of concentration per unit time.

Complete step by step answer:
The reaction given to us is

${\text{2A + B}} \to {{\text{A}}_{\text{2}}}{\text{B}}$

The rate law equation given to us for the above reaction is

${\text{rate = }}k{\text{[A][B}}{{\text{]}}^{\text{2}}}$

Where,

$k$= rate constant = \[2.0 \times {10^{ - 6}}{\text{mo}}{{\text{l}}^{{\text{ - 2}}}}{{\text{L}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}}\]

${\text{[A]}}$= concentration of reactant A

${\text{[B]}}$= concentration of reactant B

Now, calculate the initial rate of the reaction when ${\text{[A] = 0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}$and ${\text{[B] = 0}}{\text{.2 mol }}{{\text{L}}^{{\text{ - 1}}}}$

${\text{rate = }}2.0 \times {10^{ - 6}}{\text{mo}}{{\text{l}}^{{\text{ - 2}}}}{{\text{L}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}}{\text{( 0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{)(0}}{\text{.2 mol }}{{\text{L}}^{{\text{ - 1}}}}{)^{\text{2}}}$

${\text{rate = 8}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$

Thus, the rate of reaction when ${\text{[A] = 0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}$and ${\text{[B] = 0}}{\text{.2 mol }}{{\text{L}}^{{\text{ - 1}}}}$ is ${\text{ 8}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$

Now, calculate the rate of reaction after ${\text{[A]}}$ is reduced to ${\text{0}}{\text{.06 mol}}{{\text{L}}^{{\text{ - 1}}}}$

${\text{[A]}}$ is reduced to ${\text{0}}{\text{.06 mol}}{{\text{L}}^{{\text{ - 1}}}}$ from ${\text{0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}}$

So, ${\text{reacted [A] = 0}}{\text{.1 mol }}{{\text{L}}^{{\text{ - 1}}}} - {\text{0}}{\text{.06 mol }}{{\text{L}}^{{\text{ - 1}}}} = 0.04{\text{ mol }}{{\text{L}}^{{\text{ - 1}}}}$

The mole ratio of reactant A and B is 2:1

Hence, ${\text{reacted [B] = 0}}{\text{.04 mol }}{{\text{L}}^{{\text{ - 1}}}} \times \dfrac{1}{2} = 0.02{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}$

Thus, ${\text{unreacted [B] = 0}}{\text{.2 mol }}{{\text{L}}^{{\text{ - 1}}}} - {\text{0}}{\text{.02 mol }}{{\text{L}}^{{\text{ - 1}}}} = 0.18{\text{ mol }}{{\text{L}}^{{\text{ - 1}}}}$

${\text{rate = }}2.0 \times {10^{ - 6}}{\text{mo}}{{\text{l}}^{{\text{ - 2}}}}{{\text{L}}^{\text{2}}}{{\text{s}}^{{\text{ - 1}}}}{\text{( 0}}{\text{.06mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{)(0}}{\text{.18 mol }}{{\text{L}}^{{\text{ - 1}}}}{)^{\text{2}}}$

${\text{rate = 3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{\text{ }}{{\text{s}}^{{\text{ - 1}}}}$

Thus, the rate of reaction when ${\text{[A]}}$ is reduced to ${\text{0}}{\text{.06 mol}}{{\text{L}}^{{\text{ - 1}}}}$ is ${\text{3}}{\text{.9}} \times {\text{1}}{{\text{0}}^{{\text{ - 9}}}}{\text{mol }}{{\text{L}}^{{\text{ - 1}}}}{{\text{s}}^{{\text{ - 1}}}}$.

Note:
The mathematical relation between the rate of reaction and the concentration of reaction component is known as the rate law expression. The rate constant value for the given reaction will be always constant. It is independent of change in concentration of reaction components. The rate of reaction is directly proportional to the concentration of reactants in the rate law expression. The higher the concentration of reactants in the rate law expression higher is the rate of reaction. A decrease in the concentration of reactants in the rate law expression decreases the rate of reaction.