Question

At 298 K, 1.20 mol BrCl(g) is introduced into a 10.0 −L vessel, and equilibrium is established in the reaction.
BrCl(g)⇌1/2Br2(g)+1/2Cl2(g)

A) Calculate the amount of BrCl(g) present when equilibrium is established. [Hint: Use ΔG0[Br2(g)]=3.11kJ/mol, ΔG0[BrCl(g)]=−0.98kJ/mol.]

B) Calculate the amount of Br2(g) present when equilibrium is established.

C) Calculate the amount of Cl2(g) present when equilibrium is established.

Answer 1

here is a similar problem. I recommend converting concentration to moles/liter.
https://www.jiskha.com/display.cgi?id=1332563637

Answer 2

Thx. Here's what i did, but its incorrect

ΔG0 = 1/2 (3.11-(-0.98))
=2.535

2.535 = -RT ln K
2.535 x10^3 = -(8.314x298) lnK
K= 0.359

2BrCl <=> Br2+ Cl2
I 0.12 0 0
C -2x x x
E 0.12-2x x x

0.359 = x^2 / 0.12-2x
0 = x^2 + 0.718x-0.04308
x = 0.05568M

[BrCl] = 0.12-2x
= 0.12-2(0.05568M)
= 8.64x10^-3
( 8.64x10^-3 )(10L)
= 0.0864mol

[Br2] = [Cl2] = 0.05568x10L
=0.5568mol

Answer 3

I believe your error is that you wrote the rxn as BrCl ==> 1/2 Br2 + 1/2 Cl2 and solved for K for that rxn BUT you then used the concns and the same K for 2BrCl ==> Br2 + Cl2. Remember that K for the first reaction is not the same as K for the second rxn. If I were you I would go back and calculate dGo and K for 2BrCl ==> Br2 + Cl2, That should end up being (0.359)^2 then rework the problem.