Question

In the first order reaction 75% of the reactant disappeared in 1.388hrs. Calculate the rate constant of the reaction:
A. $${\text{1}}{{\text{s}}^{{\text{ - 1}}}}$$
B. $${\text{2}}{\text{.8}} \times {\text{1}}{{\text{0}}^{ - 4}}{{\text{s}}^{{\text{ - 1}}}}$$
C. $$17.2 \times {\text{1}}{{\text{0}}^{ - 3}}{{\text{s}}^{{\text{ - 1}}}}$$
D. $$1.8 \times {\text{1}}{{\text{0}}^{ - 3}}{{\text{s}}^{{\text{ - 1}}}}$$

Answer 1

Hint: The rate law is also called a rate equation is an expression which helps us to relate between the rate of the reaction and the concentration of the reactant which participates in the chemical reaction. The overall order of the reaction is defined as the sum of the partial order of the reactants expressed in the rate law expression. If the reaction is of zero order then on doubling the reactant concentration there will be no change in the reaction rate.

Complete step by step answer:
The formula which we will be using for this question is the following:
$${\text{k = }}\dfrac{{{\text{2}}{\text{.303}}}}{{\text{t}}}{\text{log}}\dfrac{{{\text{100}}}}{{{\text{100 - x}}}}$$
Here t is the time and x is the amount of substance which has been used.
So the t = 1.388hrs
So first we will convert the time in seconds:
So we need to multiply the time given by 3600 so we get,
t = $$1.388 \times 3600 = 4996.8$$seconds
so substituting the value in the formula given above :
$${\text{k = }}\dfrac{{{\text{2}}{\text{.303}}}}{{4996.8}}{\text{log}}\dfrac{{{\text{100}}}}{{{\text{100 - 75}}}}$$
so on simplifying the above equation we get,
K = $${\text{2}}{\text{.8}} \times {\text{1}}{{\text{0}}^{ - 4}}{{\text{s}}^{{\text{ - 1}}}}$$

So, the correct answer is Option B.

Note: In the first order of the reaction on doubling the concentration of reactants the rate of the reaction will be doubled. Whereas in the second order of the reaction on doubling the concentration of reactants we will see that the rate will quadruple. Whereas in the third order of reaction the rate would increase by eight times.