Question

2 NOCl(g) <=> 2 NO(g) + Cl2(g), K= 1.6E-5. Calculate the concentrations when:

a). 6.0 mol of NO and 2.0 mol of Cl2 in a 1.2-L flask

b).  2.2 mol of NOCl, 2.2 mol of NO, and 1.1 mol of Cl₂ in a 1.0-L flask

c).  1.01 mol/L concentration of all three gases

Please show work and don't just solve the top problem. I need all three and I don't know how to correctly solve either of them!!

Answer 1

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

Equilibrium constant=Keq=[NO]^2 [Cl2]/[ NOCl]^2

ICE table=

	[ NOCl]	[NO]	[Cl2]
Initial concentration	0	6.0	2.0
change	+2x	-2x	-x
Equilibrium concentration	2x	6.0-2x	2.0-x

Keq=[NO]^2 [Cl2]/[ NOCl]^2

1.6*10^-5=(6.0-2x) ^2(2.0-x)/(2x)^2 [ignore x<<<6.0 and 2.0)

1.6*10^-5=(6.0) ^2(2.0-)/(2x)^2

1.6*10^-5=(36.0)(2.0)/4x^2

x^2 =(36.0)(2.0)/4*( 1.6*10^-5)=11.25*10^-5=1.125*10^-4

x=sqrt(1.125*10^-4)=1.1*10^-2=0.011

[ NOCl]=2x=2*0.011=0.022 moles

[ NOCl]=0.022 moles/volume of flask=0.022mol/1.2L=0.0183M

[ NOCl]=0.0183M

[NO]= 6.0-2x=6.0-0.022 moles=5.98 moles

[NO]=5.98mol/1.2L=4.98M

[Cl2]=2.0-x=2.0-0.011=1.99 mol

[CL2]=1.99mol/1.2L=1.65M

b) 2 mol of NOCl, 2.2 mol of NO, and 1.1 mol of Cl2 in a 1.0-L flask

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

ICE table=

	[ NOCl]	[NO]	[Cl2]
Initial concentration	2.0	2.2	1.1
After complete reaction	0	2.2+2.0=4.2	1.1+1.0=2.1
change	+2x	-2x	-x
Equilibrium concentration	2x	4.2-2x	2.1-x

Keq=[NO]^2 [Cl2]/[ NOCl]^2

1.6*10^-5=(4.2-2x) ^2(2.1-x)/(2x)^2 [ignore x<<< 4.2 and 2.1)

1.6*10^-5=(4.2) ^2(2.1)/(2x)^2

   1.6*10^-5 =37.04/4x^2

X^2=37.04/4*(1.6*10^-5)

X^2=5.79 *10^-5

X^2=0.579*10^-4

X=0.76*10^-2=0.008 moles

[NOCl]=2x=2*0.008=0.016 mol or 0.016mol1L=0.016M

[NO]=4.2+2x=4.2-0.016=4.18 mol or 4.18mol/1L=4.18M

[cl2]=2.1-x=2.1-0.008=2.09 mol or 2.09M

c)same asb