Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

Read the passage given below and answer based it.

The rate of a reaction is concerned with decrease in concentration of reactants or increase in concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst, affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. Molecularity is defined only for an elementary reaction. Molecularity and order of an elementary reaction are same.

Which of the following expressions is correct for the rate of the reaction given below?

\(H_2O_2 (aq) + 3I^- (aq) + 2H^+ \longrightarrow 2H_2O (l) + I_3^-\)

Options:

\(\frac{\Delta [I^-]}{\Delta t} = 3\frac{[\Delta H^+]}{\Delta t}\)

\(\frac{\Delta [I^-]}{\Delta t} = \frac{2}{3}\frac{[\Delta H^+]}{\Delta t}\)

\(\frac{\Delta [I^-]}{\Delta t} = \frac{3}{2}\frac{[\Delta H^+]}{\Delta t}\)

\(\frac{\Delta [I^-]}{\Delta t} = 2\frac{[\Delta H^+]}{\Delta t}\)

Correct Answer:

\(\frac{\Delta [I^-]}{\Delta t} = \frac{3}{2}\frac{[\Delta H^+]}{\Delta t}\)

Explanation:

The correct answer is option 3. \(\frac{\Delta [I^-]}{\Delta t} = \frac{3}{2}\frac{[\Delta H^+]}{\Delta t}\).

The given reaction is

\(H_2O_2 (aq) + 3I^- (aq) + 2H^+ \longrightarrow 2H_2O (l) + I_3^-\)

\(\text{Rate = }-\frac{\Delta [H_2O_2]}{\Delta t} = -\frac{1}{3}\frac{\Delta [I^-]}{\Delta t} = -\frac{1}{2}\frac{\Delta [H^+]}{\Delta t} = \frac{1}{2}\frac{\Delta [H_2O]}{\Delta t} = \frac{\Delta [I_3^-]}{\Delta t}\)

or, we can write

\(\frac{\Delta [I^-]}{\Delta t} = \frac{3}{2}\frac{\Delta [H^+]}{\Delta t}\)