Class 12 Chemistry Chemical Kinetics Factors affecting rate of reaction

Factors affecting rate of reaction

Nature of reactant

Nature of bonding in the reactants determines the rate of a reaction. The ionic compounds react faster compared to covalent compounds due to requirement of energy in covalent compounds to cleave the existing binds. 

The reaction between ionic compounds:

Precipitation of AgCl

AgNO3 + NaCl --> AgCl + NaNO3

The reactions between covalent compounds:

Temperature

Rate of reaction increases with the rise in temperature due to increase in average kinetic energy which in turn increases the number of molecules having greater energy than threshold energy and consequently increasing the number of effective collisions. The rate of a reaction is doubled (i.e., increased by 100%) with 10 oC rise in temperature.

Pressure

Increase in partial pressure increases the number of collisions. Therefore, the rate of reactions involving gaseous reactants increases with the increase in partial pressures.

Catalyst

 A catalyst increases the rate of reaction by giving an alternative path with lower activation energy (Ea’) for the reaction to proceed. 

Concentration of reactants

Increase in concentration increases the number of collisions and the activated collisions between the reactant molecules. According to the collision theory, rate is directly proportional to the collision frequency. Consequently, the rate of a reaction increases with the rise in the concentration of reactant.

Surface area

The rate of a reaction increases with increase in the surface area of solid reactant.

 

PROBLEM.   For the reaction: 2A + B → A2B  , The rate = k[A][B]2with k= 2.0 x 10-6mol-2L2s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L-1.

SOLUTION.   Rate = k [A][B]2

= (2.0 × 10 - 6mol - 2L2s - 1) (0.1 mol L - 1) (0.2 mol L - 1)2

= 8.0 × 10 - 9mol - 2L2s - 1

Reduction of [A] from 0.1 mol L - 1to 0.06 mol – 1

The concentration of A reacted = (0.1 - 0.06) mol L - 1 = 0.04 mol L - 1

The concentration of B reacted= 1/2 x 0.04 mol L-1 = 0.02 mol L - 1

The concentration of B available, [B] = (0.2 - 0.02) mol L - 1

= 0.18 mol L - 1

After reduction of [A] to 0.06 mol L - 1

The rate of the reaction

Rate = k [A][B]2

= (2.0 × 10 - 6mol - 2L2s - 1) (0.06 mol L - 1) (0.18 mol L - 1)2

= 3.89 mol L - 1s - 1

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