A reaction has a standard free-energy change of –4.60 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?

The problem doesn't tell us what the reaction is; i.e., is this K for

A + B ==> C or for
C ==> A + B.
I am assuming it is for A + B ==> C
Use dG = -RT*lnK and solve for Kc.
Then
.............A + B ==> C
I..........0.30.0.40....0
C...........-x...-x.....x
E........0.3-x..0.4-x...x
Then substitute the E line into the Kc expression and solve for x, then evaluate the other terms.

To determine the concentrations of A, B, and C at equilibrium, we need to use the relationship between the standard free-energy change (∆G°), the equilibrium constant (K), and the concentrations of the reactants and products.

The equilibrium constant (K) is related to the standard free-energy change (∆G°) through the equation:

∆G° = -RT ln(K)

Where:
∆G° is the standard free-energy change (-4.60 kJ/mol in this case),
R is the gas constant (8.314 J/mol∙K),
T is the temperature in Kelvin (25 + 273.15 = 298.15 K),
ln denotes the natural logarithm, and
K is the equilibrium constant.

First, we need to convert the given ∆G° from kJ to J:
∆G° = -4.60 kJ/mol * 1000 J/kJ = -4600 J/mol

Next, using the equation with the known values:
-4600 J/mol = -(8.314 J/mol∙K) * (298.15 K) * ln(K)

Now, we can solve the equation for ln(K):

ln(K) = -4600 J/mol / [(8.314 J/mol∙K) * (298.15 K)]
ln(K) ≈ -2.261

Using the relationship between the equilibrium constant (K) and the concentrations of reactants and products, we have the following equation for the reaction:

A + B ⇌ C

K = [C] / ([A] * [B])

Let's denote the concentrations of A, B, and C at equilibrium as [A]eq, [B]eq, and [C]eq, respectively.

Given that at the beginning of the reaction, [A] = 0.30 M, [B] = 0.40 M, and [C] = 0 M, and assuming they change by x M at equilibrium:

[A]eq = (0.30 - x) M
[B]eq = (0.40 - x) M
[C]eq = x M

We can substitute these values into the equilibrium constant expression:

K = [C]eq / ([A]eq * [B]eq)
K = x / [(0.30 - x) * (0.40 - x)]

Now, we can solve for x.

Let's substitute ln(K) ≈ -2.261 and solve for x:

-2.261 = -4600 J/mol / [(8.314 J/mol∙K) * (298.15 K)]
-2.261 ≈ ln[(x / [(0.30 - x) * (0.40 - x)])]

To solve this equation, we can use logarithmic properties. Taking the exponential of both sides:

e^(-2.261) ≈ x / [(0.30 - x) * (0.40 - x)]

The left side of the equation is approximately 0.104.

0.104 ≈ x / [(0.30 - x) * (0.40 - x)]

Rearranging the equation, we have:

0.104 * [(0.30 - x) * (0.40 - x)] ≈ x

Expanding and simplifying:

0.1248 - 0.416x + 0.4x^2 - 0.13x^2 ≈ x

Combining like terms:

0.27x^2 - 0.416x + 0.1248 ≈ x

Rearranging the equation:

0.27x^2 - 1.416x + 0.1248 ≈ 0

At this point, you can solve this quadratic equation using the quadratic formula or numerical methods to approximate the value of x. Once you find the value of x, you can substitute it back in the expressions for [A]eq, [B]eq, and [C]eq to find their concentrations at equilibrium.