Chapter 17, Problem 71A

Interpretation Introduction

Interpretation:

The equilibrium constant of the given reaction have to be determined.

Concept Introduction:

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

  $aA+bB\rightleftharpoons cC+dD$

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

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

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

  $Atequilibriumtherateoftheforwardreaction=therateofthebackwardreaction$

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

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

  $K_{eq}istheequilibriumconstantofthereaction.$

Answer to Problem 71A

The equilibrium constant of the given reaction is 1.0.

Explanation of Solution

Data given: At initially [N₂] = 5.0 M and [H₂] = 4.0 M and at equilibrium [H₂] = 1.0 M

The given chemical equation is-

  $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$

  $Whent=0,[N_2]=5.0Mand[H_2]=4.0M$

  $Atequilibrium,[N_2]=(5.0-x)M,[H_2]=(4.0-3x)Mand[N{H}_3]=2xM$

  $Giventhat,[H_2]=(4.0-3x)M=1.0M$

  $So,thevalueofx=1$

  $Atequilibrium,[N_2]=4.0M,[H_2]=1.0Mand[N{H}_3]=2.0M$

  $K_{eq}=\frac{{[N{H}_3]}^2}{[N_2]{[H_2]}^3}$

  $=\frac{{(2.0)}^2}{(4.0)\times {(1.0)}^3}=1.0$