Question

In a reaction $2{\text{A}} \to {\text{products}}$ , the concentration of A decreases from $0.5{\text{mol}}{{\text{L}}^{ - 1}}$ to $0.4{\text{mol}}{{\text{L}}^{ - 1}}$ in 10 minutes. Calculate the rate during this interval.

Answer 1

Rate of reaction is equal to change in concentration of reactant or product with respect to interval of time and divide by their respective stoichiometric coefficient. It is also the instantaneous rate of formation of product in the reaction.

Complete step by step answer:
Rate of a reaction is defined as the rate of change of concentration of any of the reactants or the product in a given time. Rate of reaction with respect to reactant is defined as the rate at which reactant is converted into product. Rate of reaction in terms of reactant can be represented as:
${\text{rate of reaction}} = - \dfrac{{{\text{change in concentration of reactant}}}}{{{\text{time interval}}}}$ ,
A negative sign is used to show the concentration of reactant keeps on decreasing with time and as rate of reaction cannot be a negative value. Minimum value of rate of reaction is zero when the reaction is stopped. Similarly, rate of reaction in terms of product is written as:
${\text{rate of reaction}} = \dfrac{{{\text{change in concentration of product}}}}{{{\text{time interval}}}}$ ,
which indicates concentration of product increases with time.
For example: ${\text{mA}} + {\text{nB}} \to {\text{xC}} + {\text{yD}}$ reaction is given in which m moles of A react with n moles of B to give x mole of C and y mole of D. For this reaction rate or rate of reaction can be given as: ${\text{rate}} = - \dfrac{1}{{\text{m}}}\dfrac{{{\text{d}}\left[ {\text{A}} \right]}}{{{\text{dt}}}} = - \dfrac{1}{{\text{n}}}\dfrac{{{\text{d}}\left[ {\text{B}} \right]}}{{{\text{dt}}}} = \dfrac{1}{{\text{x}}}\dfrac{{{\text{d}}\left[ {\text{C}} \right]}}{{{\text{dt}}}} = \dfrac{1}{{\text{y}}}\dfrac{{{\text{d}}\left[ {\text{D}} \right]}}{{{\text{dt}}}}$ .
Such rate of reaction is known as instantaneous rate of reaction.
Now as given in question, $2{\text{A}} \to {\text{products}}$ , , the concentration of A decreases from $0.5{\text{mol}}{{\text{L}}^{ - 1}}$ to $0.4{\text{mol}}{{\text{L}}^{ - 1}}$ in 10 minutes. Rate of this reaction can be calculated as: ${\text{rate}} = - \dfrac{1}{2}\dfrac{{{{\left[ {\text{A}} \right]}_{{\text{final}}}} - {{\left[ {\text{A}} \right]}_{{\text{initial}}}}}}{{{\text{time interval}}}}$ . By putting value we get: ${\text{rate}} = - \dfrac{1}{2}\dfrac{{0.4 - 0.5}}{{10}} = 0.005{\text{mol}}{{\text{L}}^{ - 1}}{\text{mi}}{{\text{n}}^{ - 1}}$ .

Thus, the correct answer is $0.005{\text{mol}}{{\text{L}}^{ - 1}}{\text{mi}}{{\text{n}}^{ - 1}}$ .
Note:
The order of the reaction with respect to some component is the power to which the concentration of the components is raised in rate law. Order is found by experiment only it cannot be deduced from the equation for a given balanced reaction.