Standard Gibbs energy from equilibrium pressure constant? This is the question:
ID: 553619 • Letter: S
Question
Standard Gibbs energy from equilibrium pressure constant?
This is the question:
The answer is meant to be B. I have tried using -RTln(K) and I get answer A. I also tried converting kp to kc but that didn't work either so I am looking for someone to walk me through how to do this.
A18. The equilibrium constant, Kp, for the reaction shown below at 25.0 °C is 9.18 x 1022. Calculate the change in standard Gibbs energy for this reaction C(s) + CO2(g) 2CO(g) (A) +1.2 x 105 kJ (B) +1.2×102 kJ (C) -1.2×105 kJ (D) +10% 104 J (E) -2.4 x 103Explanation / Answer
T= 25.0 oC
= (25.0+273) K
= 298 K
delta n = number of gaseous molecule in product - number of gaseous molecule in reactant
delta n = 1
Kp= Kc (RT)^deltan
9.18E-22 = Kc *(0.0821*298.0)^(1)
Kc = 3.752*10^-23
we have below equation to be used:
deltaG = -R*T*ln Kc
= - 8.314*298* ln(3.752*10^-23)
= 127935 J
= 128 KJ
= 1.2*10^2 KJ
Answer: B
Feel free to comment below if you have any doubts or if this answer do not work
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