N2 + 3H2 --> 2NH3
At 25°C (delta)Ho = -92.22 kJ and (delta)So = -198.53 J/K. Using this information, calculate the equilibrium constant for the reaction at 2.43x10^2°C. (R = 8.314 J/K)
Apparently the way I'm doing it is wrong:
I divide 198.53 by 1000 then multiply by 298 to get delta(S). Then I subtract this from delta(H) to get delta(G). Then I multiply 516 with .008314... I take this answer and divide delta(G) by it, then multiply by -1. I take this to the e power... Then I use the equation Kc = (RT)^-n * Kp to get the equilibrium constant. However my answer is still wrong.
To calculate the equilibrium constant (K) for the given reaction at 2.43x10^2°C, you need to use the equation:
ΔG = ΔH - TΔS (where ΔG is the Gibbs free energy change)
First, let's convert the temperature from 2.43x10^2°C to Kelvin:
T (in Kelvin) = 2.43x10^2 + 273 = 473K
Then, let's convert the standard entropy change (ΔS) from J/K to kJ/K:
ΔS = -198.53 J/K * (1 kJ / 1000 J) = -0.19853 kJ/K
Now, substitute the values into the equation:
ΔG = -92.22 kJ - (473K * (-0.19853 kJ/K))
= -92.22 kJ + 93.99 kJ
= 1.77 kJ
Next, convert the gas constant (R) from J/K to kJ/K:
R = 8.314 J/K * (1 kJ / 1000 J)
≈ 0.008314 kJ/K
Now, calculate e^(ΔG / (RT)):
e^(1.77 kJ / (0.008314 kJ/K * 473K))
≈ e^(2.28)
Finally, to calculate the equilibrium constant (Kc), use the equation:
Kc = (RT)^(-Δn) * e^(ΔG / (RT))
Since the stoichiometric coefficient of NH3 is 2 (the change in the number of moles for the reaction is Δn = 2 - (1 + 3) = -2), we can substitute the values:
Kc = (0.008314 kJ/K * 473K)^2 * e^(1.77 kJ / (0.008314 kJ/K * 473K))
≈ 4.38 * 10^3
Therefore, the equilibrium constant (Kc) for the reaction at 2.43x10^2°C is approximately 4.38 * 10^3.