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.
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
1 answer
See my response below at your first answer from Bob Pursley.