Question

At 35 degrees C, Kc = 1.6x10^-5 for the reaction:

2NOCl(g),<--> 2NO(g) + Cl2(g)

Calculate the concentrations of all species at equilibrium for each of the following origional mixtures.

a. 1.00 mol NOCl in a 1.0 L container

b. 1.00 mol NO and 0.50 mol Cl2 in a 1.0 L container

c. 1.00 mol NO and 0.75 mol Cl2 in a 1.0 L container

The answers are: a. [NOCl]= 1.0M, [NO]= 0.032M, [Cl2]= 0.016M; b. [NOCl]= 1.0M, [NO]= 0.032M, [Cl2]= 0.016M; c. [NOCl]= 1.0M, [NO]= 0.00080M, [Cl2]= 0.25M

Please show work!!

Answer 1

For the 1^st case, Concentration of NOCl= 1/1=1M

The reaction is 2NOCl(g) <------>2NO(g)+ Cl2(g)

K= [Cl2] [NO]²/ [NOCl]² = 1.6*10^-5

let x= concentration of Cl2 at equilibrium

at Equilibrium [NOCl]= 1-2x, [NOCl]=2x and [Cl2]=x

hence   x*(2x)²/ (1-2x)²= 1.6*10^-5,

x³/(1-2x)⁵= 4*10^-6, when solved using excel, x= 0.016

at Equilibrium [NOCl]= 1-2*0.016= 0.97M, [Cl2]= 0.016 and [NOCl]=2*0.016=0.032M

2. for this case [No]= 1/1=1M and [Cl2]=0.5/1=0.5M

Q= reaction coefficient = [NO] [Cl2]²/[NOCl]²= 1*0.5*0.5/0= infinity

so the Q>>K,   the backward reaction takes place

let x= drop in concentration of [Cl2] to reach equilibrium

at Equilibrium [Cl2]=0.5-x, [NO] =1-2x, [NOCl]=2x

KC= (0.5-x)*(1-2x)²/ (2x)²= 1.6*10^-5,

(0.5-x)*(1-2x)²/x²= 1.6*4*10^-5 = 6.4*10^-5,when solved using excel, x= 0.4845

hence at [Cl2]=0.5-0.4845=0.0155M, [NO] =1-2*0.4845= 0.031 and [NOCl]= 2*0.4845= 0.969M

for the third case initially [No]= 1/1=1M, [Cl2]=0.75/1= 0.75M

here also back ward reaction takes place and [NOCl] forms

if x= drop in concentration of Cl2 to reach equiirbium, at Equilibrium [NO]= 1-2x, [Cl2]=0.75-x, [NOCl] =2x

hence (0.75-x)*(1-2x)²/ x²= 6.4*10^-5

^when sovled ising excel, x= 0.49599, [NOCl]= 2*0.49599 =0.99M, [NO]= 1-2*0.49599 =0.008M, [Cl2]=0.75-0.49599 =0.254M