I. (12 pts.) A chemical reaction 3 A (g) = B (g) with G\'-+34.2 kJ (at 298 K). a

Question

I. (12 pts.) A chemical reaction 3 A (g) = B (g) with G'-+34.2 kJ (at 298 K). a. (8) Calculate Keq for the equilibrium. b. (4) Is it possible to predict the sign of AH for this reaction? If so, what is the sign? If not, why not? In all cases, justify your answer. 2. (12 pts.) Find @for the reaction below, given the following reactions and G values: C2H2 (g) + 5/2O2 (g) 2 CO2 (g) + H2O (g) C2H6 (g) C2H2 (g) + 2 H2 (g) H2 (g) + ½ O2 (g) H2O (g) 2 CO2 (g) + 3 H2O (g) C2H6 (g) + 7/2 O2 (g) G = 283.5 kJ G =-213.7 kJ G = +849 kJ

Explanation / Answer

1)

1)

a)

T = 298 K

deltaG = 34.2 KJ/mol

deltaG = 34200 J/mol

we have below equation to be used:

deltaG = -R*T*ln Kc

34200 = - 8.314*298.0* ln(Kc)

ln Kc = -13.8038

Kc = 1.01*10^-6

Answer: 1.01*10^-6

b)

Here:

delta G is positive

since reactant has more gases, delta S is negative

we know:

delta G = delta H - T*delta S

since delta G is positive and delta S is negative, delta H can be og any sign

So, it is not possible to predict the value

Only 1 question at a time please

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