Question

∆G°=-139kJ/mol. 2SO2(g) + O2(g) → 2SO3(g) Calculate the Kp at 25 oC for the reaction Are...

∆G°=-139kJ/mol.

2SO2(g) + O2(g) → 2SO3(g)

Calculate the Kp at 25 oC for the reaction

Are products or reactants favored at equilibrium? Why?

Homework Answers

Answer #1

T= 25.0 oC

= (25.0+273) K

= 298 K

G = -139 KJ/mol

G = -139000 J/mol

we have below equation to be used:

deltaG = -R*T*ln Kc

-139000 = - 8.314*298.0* ln(Kc)

ln Kc = 56.1033

Kc = 2.319*10^24

delta n = number of gaseous molecule in product - number of gaseous molecule in reactant

delta n = -1

Kp= Kc (RT)^delta n

Kp = 2.319*10^24*(0.0821*298.0)^(-1)

Kp = 9.5*10^22

Answer: 9.5*10^22

Since Kp>>1, it is produce favoured

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