1 answer

Suppose the reaction A <---> B in the forward direction is first order in A and...

Question:

Suppose the reaction

A <---> B

in the forward direction is first order in A and the rate constant is 1.50 x 10-2s-1. The reverse reaction is first order in B and the rate constant is 4.50 x 10-2s-1 at the same temperature. What is the value of the equilibrium constant for the reaction

A <---> B

at this temperture?


Answers

Concepts and reason

Chemical equilibrium: In a chemical reaction, when the rate of the forward reaction is equal to the rate of the backward reaction, equilibrium is said to have been attained. The concentration of the reactant and product were maintained in the chemical equilibrium.

Equilibrium constant in terms of (к.) : Equilibrium constant is defined as the ratio of the concentration of the product and the concentrations of reactants, raised to their respective stoichiometric powers.

For the gaseous reactant and product the equilibrium can be expressed in terms of partial pressure, Then equilibrium constant will be (Кр) .

Reaction quotient: Reaction quotient is the ratio of the initial concentrations of reactants to the initial concentrations of products.

Fundamentals

Equilibrium constant: Equilibrium constant is the ratio of the rate constant of a forward reaction to the rate constant of a reverse reaction.

Let the reaction be as follows, A В

The net equilibrium constant of the reaction K can be written as

K [в] [А]

Given chemical reaction:

A В

Given data:

Forward rate constant, k, =1.50x 102s1 Reverse rate constant, k, = 4.50x 102s1

Substituting the respective values, 1.50x 102s K 4.50x 10 s K=3.33x10

Ans:

Hence, the numerical value of the equilibrium constant for the given equilibrium reaction, A В at particular temperature is3.33x10 1 .

.

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