Chapter 17, Problem 64A

Interpretation Introduction

Interpretation:

A quantitative molecular picture that shows the system reaches equilibrium have to be drawn and have to be given support with calculations.

Concept Introduction:

Chemical equilibrium: Chemical equilibrium is a state of a chemical reaction when the rate of forward reaction is equal to the rate of backward reaction. The chemical reaction follows the law of mass action. The law of mass action states that the rate of a chemical reaction is directly proportional to the product of active masses of the reactants.

Answer to Problem 64A

The quantitative molecular picture is-

  $aA+bB\rightleftharpoons cC+dD$

  $Rateoftheforwardreaction{r}_f=k_f{[A]}^a{[B]}^b$

  $Rateofthebackwardreaction{r}_b=k_b{[C]}^c{[D]}^d$

  $k_{f}and{k}_{b}aretherateconstantsofforwardandbackwardreactionrespectively$

  $Atequilibriumtherateoftheforwardreaction=therateofthebackwardreaction$

  $k_f{[A]}^a{[B]}^b=k_b{[C]}^c{[D]}^d$

  $\frac{k_f}{k_b}=K_{eq}=\frac{{[C]}^c{[D]}^d}{{[A]}^a{[B]}^b}$

  $K_{eq}istheequilibriumconstantofthereaction.$

  

The calculated equilibrium constant K_eq value matches with the given data.

Explanation of Solution

Data given: K_eq = 25.0

  

The chemical equation is-

  

  $\begin{array}{l}A(g)+B(g)\rightleftharpoons C(g)+D(g)\\ K_{eq}=\frac{[C][D]}{[A][B]}\end{array}$

Initially 12 molecules of A and 12 molecules of B are present in the container.

If 10 molecules of A react with 10 molecules of B:

The container will contain 2 molecules of A, 2 molecules of B, 10 molecules of C and 10 molecules of D.

  $\begin{array}{l}{K}_{eq}=\frac{[C][D]}{[A][B]}\\ =\frac{10.0\times 10.0}{2.0\times 2.0}=25.0\end{array}$

The calculated equilibrium constant K_eq value matches with the given data.