A reaction A+B---> C has a standard free energy change of -5.13 kj/mol at 25 degrees celcius, What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction there concentrations are .3M, .4M, and 0M respectively?

I would use DG = -RTlnK and solve for K. Then set up an ICE chart for the equilibrium and solve for each concn.

To determine the concentrations of A, B, and C at equilibrium, we need to use the equation for the standard free energy change:

ΔG° = -RT ln(K)

where ΔG° is the standard free energy change, R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant.

First, let's convert the temperature from Celsius to Kelvin by adding 273.15:

T = 25°C + 273.15 = 298.15 K

Next, we can rearrange the equation to solve for K:

K = e^(-ΔG° / RT)

Now, substitute the values in:

K = e^(-(-5.13 kJ/mol) / (8.314 J/(mol·K) * 298.15 K)

Since the standard units for ΔG° are in kJ/mol and R is in J/(mol·K), we need to convert ΔG° to J/mol by multiplying -5.13 kJ/mol by 1000:

K = e^(5130 J/mol / (8.314 J/(mol·K) * 298.15 K)

After simplifying the equation, we get:

K = e^(-20.70)

Now we can find the concentrations of A, B, and C at equilibrium using the equilibrium constant expression:

K = [C]^c / ([A]^a * [B]^b)

The reaction A + B → C shows a 1:1 stoichiometric ratio between A, B, and C. So the equilibrium constant expression becomes:

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

We can now substitute the values provided:

[C] / ([A] * [B]) = e^(-20.70)

At the beginning of the reaction, the concentrations are given as:

[A] = 0.3 M
[B] = 0.4 M
[C] = 0 M

Substituting these values into the equation, we get:

0 / (0.3 * 0.4) = e^(-20.70)

Since the denominator is zero, this means that the reaction has not yet reached equilibrium. Therefore, we cannot determine the concentrations of A, B, and C at equilibrium based on the given information.