Question

The reaction

2NOCl(g)=2NO(g)+Cl₂(g)

comes to equilibrium at 1 bar total pressure and 227C when the partial pressure of the nitrosyl chloride, NOCl, is 0.64 bar. Only NOCl was present initially. (a) Calculate delta_rG^o for this reaction. (b) At what total pressure will the partial pressure of Cl₂ be at 0.1 bar?

Answer 1

	NOCl	NO	Cl2	Total
Initial	n0	0	0	n0
equ	n0 − 2ξ	2ξ	ξ	n0 + ξ

Here ξ goes from 0 to n0/2. It is more convenient to work with reduced variable ξ ′ = ξ/n0. In this case the above table takes the form:

	NOCl	NO	Cl2	Total
Initial	1	0	0	1
equ	1 − 2ξ ′	2ξ ′	ξ ′	1 + ξ

both ξ and ξ would give the same molar fractions and hence the same equilibrium constant K.

At equilibrium Ptot = 1 bar and PNOCl = 0.64 bar

. Because PNOCl = yNOClPtot, we get

yNOCl = 0.64.

Above we have an expression that relates yNOCl and ξ ′ to each other

yNOCl =1 − 2ξ ′ /1 + ξ ′ eq =0.64

ξ ′ eq = 0.136

K= (pNO)^2 (Pcl2) /(pNOcl)^2

plugging value

= 0.0167

∆rG ◦ = −RT ln(K) = − 8.314JK^−1mol^−1 × (500 K) ln (0.0167)

= 17.0 kJ mol-----answer

===========

K = (y eq NO)^ 2 ( y eq Cl2 ) / (y eq NOCl)^ 2 × ( Ptot /P◦)

= 0.0167

= (y eq NO) ^2 × (0.1 bar /1 bar) /(y eq NOCl)^ 2 == 0.0167 =

y eq NO / y eq NOCl = square root of 0.0167

2ξ ′ eq/ 1 − 2ξ ′ = 0.409 ⇒ ξ ′ eq = 0.145

y eq Cl2 = 0.145 1 /( + 0.145)= 0.127

P tot = 0.787 bar----answer