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
See my response below at your first answer from Bob Pursley.
To calculate the amounts of each species at equilibrium, you need to use the equilibrium constant expression and the balanced equation for the reaction.
A) The equilibrium constant expression for the reaction is given by:
K = ([Br2]^(1/2) * [Cl2]^(1/2)) / [BrCl]^2
We can calculate the value of K using the equation:
ΔG0 = -RT ln K
Given:
ΔG0[Br2(g)] = 3.11 kJ/mol
ΔG0[BrCl(g)] = -0.98 kJ/mol
ΔG0 = 1/2 (ΔG0[Br2(g)] - ΔG0[BrCl(g)])
ΔG0 = 1/2 (3.11 - (-0.98)) kJ/mol
ΔG0 = 2.535 kJ/mol
Now, we can use the equation to calculate K:
2.535 kJ/mol = -(8.314 J/(mol*K) * 298 K) * ln K
2.535 * 10^3 = -2474.372 * ln K
ln K = -2.535 * 10^3 / -2474.372
ln K = 1.0255
K = e^(ln K)
K = e^(1.0255)
K ≈ 2.782
Now, we can calculate the amount of BrCl(g) present when equilibrium is established:
[BrCl] = [Initial BrCl] - 2x
[BrCl] = 1.20 mol - 2(0.05568 mol)
[BrCl] ≈ 0.0864 mol
B) Since the stoichiometric coefficient of Br2 is 1/2 in the balanced equation, the amount of Br2(g) present at equilibrium will be equal to 1/2 of the amount of BrCl(g):
[Br2] = (1/2) * [BrCl]
[Br2] ≈ 0.0864 mol
C) Similarly, the amount of Cl2(g) present at equilibrium will also be equal to 1/2 of the amount of BrCl(g):
[Cl2] = (1/2) * [BrCl]
[Cl2] ≈ 0.0864 mol
Therefore, at equilibrium, the amount of BrCl(g) present is approximately 0.0864 mol, the amount of Br2(g) present is approximately 0.0864 mol, and the amount of Cl2(g) present is approximately 0.0864 mol.