Help me set up this problem. I have been looking at a few examples of similar problems and I keep getting the wrong answers. I'm getting really confused.

Q: The standard free energy change ΔG∘ and the equilibrium constant K for a reaction can be related by the following equation:

ΔG∘=−RTlnK
where T is the Kelvin temperature and R is equal to 8.314J/(mol⋅K).

a. Calculate the equilibrium constant for the reaction forming nitric oxide at room temperature, 25 ∘C.
Express your answer numerically.

A: What I got was 1.93*10^-71, which is wrong. I can't figure out where I went wrong.

Show the equation you used. I don't know the one you used.

Show the work you did.
I'll find the error.

The standard free energy change, ΔG∘, and the equilibrium constant K for a reaction can be related by the following equation:

ΔG∘=−RTlnK
where T is the Kelvin temperature and R is equal to 8.314 J/(mol⋅K).

Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C:

N2(g)+O2(g)→2NO(g)

I'm also having trouble with this question, my incorrect answer is 1.94x10^-31

ΔH= 182.6 kJ/mol
ΔS= 2.48x10^-2 kJ/mol*K
ΔG= 1.73x10^2 kJ/mol*K

The equation I thought I was suppose to use is this one.
K = e^(-ΔG/RT)

Thanks in advance for your help. :)

To calculate the equilibrium constant (K) for the reaction forming nitric oxide at room temperature (25°C), we first need to convert the temperature from Celsius to Kelvin:

T (Kelvin) = T (Celsius) + 273.15
T (Kelvin) = 25 + 273.15 = 298.15 K

Now we can plug the values into the equation ΔG∘ = -RTlnK:

ΔG∘ = -RTlnK

where ΔG∘ is the standard free energy change (-), R is the gas constant (8.314 J/(mol⋅K)), T is the temperature in Kelvin (298.15 K), and ln denotes the natural logarithm.

To solve for K, we rearrange the equation:

ΔG∘ = -RTlnK
-lnK = ΔG∘ / RT
lnK = -ΔG∘ / RT
K = e^(-ΔG∘ / RT)

Now we can substitute the given values:

K = e^(-ΔG∘ / RT)
K = e^(-ΔG∘ / (8.314 J/(mol⋅K) * 298.15 K))

To calculate K, we need to know the value of ΔG∘ for the reaction forming nitric oxide. Please provide the value of ΔG∘ to proceed with the calculation.

To help you set up the problem correctly, let's go through the steps together:

Step 1: Convert temperature to Kelvin. The given temperature is in Celsius, so we need to convert it to Kelvin.

T = 25°C + 273.15 = 298.15 K

Step 2: Plug the values into the equation ΔG∘ = -RT ln K.

ΔG∘ = -RT ln K

Step 3: Substitute the values for R and T.

ΔG∘ = -(8.314 J/(mol⋅K)) * (298.15 K) * ln K

Step 4: Solve for K.

To solve for K, we need to rearrange the equation and isolate K.

Divide both sides of the equation by -RT:

(ΔG∘) / (-RT) = ln K

Exponentiate both sides of the equation:

e^((ΔG∘) / (-RT)) = e^(ln K)

K = e^((ΔG∘) / (-RT)) = e^(-(ΔG∘) / (RT))

Step 5: Calculate K using the obtained values.

Now we can substitute the given values for ΔG∘ and the converted temperature T to calculate the equilibrium constant K.

K = e^(-(ΔG∘) / (RT))

Please note that for accurate calculations, it would be helpful if you provide the value of ΔG∘ for forming nitric oxide at room temperature in Joules per mole (J/mol).