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.
here is a similar problem. I recommend converting concentration to moles/liter.
https://www.jiskha.com/display.cgi?id=1332563637
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
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.
To find the amount of each component when equilibrium is established, we can use the Gibbs free energy equation and the equilibrium constant expression. Let's go step by step.
A) To calculate the amount of BrCl(g) present at equilibrium, we need to use the Gibbs free energy change for the reaction. The formula is:
ΔG = ΔG° + RTln(Q)
Where:
ΔG is the Gibbs free energy change,
ΔG° is the standard Gibbs free energy change,
R is the gas constant (8.314 J/mol·K),
T is the temperature in Kelvin,
Q is the reaction quotient.
From the hint given, we have:
ΔG°[Br2(g)] = 3.11 kJ/mol
ΔG°[BrCl(g)] = -0.98 kJ/mol
Let's use the reaction quotient Q:
Q = [Br2(g)]^[1/2] · [Cl2(g)]^[1/2] / [BrCl(g)]
At equilibrium, Q becomes the equilibrium constant K.
K = [Br2(g)]^[1/2] · [Cl2(g)]^[1/2] / [BrCl(g)]
Now we substitute the given values:
K = ([Br2(g)]^[1/2] · [Cl2(g)]^[1/2]) / [BrCl(g)] = 10^(-0.98 kJ/mol / (8.314 J/mol·K × 298 K))
We can then solve for [BrCl(g)] at equilibrium:
[BrCl(g)] = [Br2(g)]^[1/2] · [Cl2(g)]^[1/2] / K
B) We can solve for [Br2(g)] at equilibrium using the same approach. The equilibrium constant expression becomes:
K = ([Br2(g)]^[1/2] · [Cl2(g)]^[1/2]) / [BrCl(g)]
Rearranging the equation, we have:
[Br2(g)] = [BrCl(g)] × K / [Cl2(g)]^[1/2]
C) Similarly, we can solve for [Cl2(g)] at equilibrium using the equilibrium constant expression:
K = ([Br2(g)]^[1/2] · [Cl2(g)]^[1/2]) / [BrCl(g)]
Rearranging the equation, we have:
[Cl2(g)] = ([Br2(g)] × [BrCl(g)])^2 / K
To calculate the amounts of BrCl(g), Br2(g), and Cl2(g), substitute the known values and solve for each value using the equations derived above.