At 35 degrees C, Kc = 1.6x10^-5 for the reaction: 2NOCl(g),<--> 2NO(g) + Cl2(g)
ID: 592780 • Letter: A
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!!
Explanation / Answer
For the 1st 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
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