Determine the equilibrium constant for the following reaction.
2 Fe3+(aq) + H2(g) 2 Fe2+(aq) + 2 H+(aq)
Here's my work, but it wasn't the correct answer:
The two half-reactions:
Fe3+ + e- --> Fe2+ E = 0.77 V
2H+ + 2e- --> H2 E = 0 V
You multiply the first one by 2, so it looks like the formula above: 2 Fe3+(aq) + H2(g) 2 Fe2+(aq) + 2 H+(aq).
Ecell = 0.77 V
lnKsp = (2)(96485)(0.77) / (8.314)(298) = 1.1e26
I have a feeling the n (2 in the equation above) is incorrect, but I don't know why. I multiplied the half-reaction of Fe by 2.
Note that this is ln K and not ln Ksp.
nEF = RTlnK
Your work looks ok to me.
I think you should flip one of them Cause rn you have 4 e on the same side instead of then being able to cancel out
To determine the equilibrium constant for the given reaction, you need to set up and solve an expression using the concentrations of the species involved.
The equilibrium constant (K) is a measure of the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. For a reaction in the form:
a A + b B c C + d D
The equilibrium constant expression is:
K = [C]^c [D]^d / [A]^a [B]^b
In the given reaction, the stoichiometric coefficients (a, b, c, d) are:
a = 2 (for Fe3+)
b = 1 (for H2)
c = 2 (for Fe2+)
d = 2 (for H+)
Therefore, the equilibrium constant expression for the given reaction is:
K = [Fe2+]^2 [H+]^2 / [Fe3+]^2 [H2]
Now, you need to determine the concentrations of the involved species at equilibrium. This information is not provided in the question, so you need additional data or assumptions.
Once you have the concentrations, substitute them into the equilibrium constant expression, and solve to find the value of K.