Question

1. Keq = 8.63x10^-4 for a reaction. Which of the following must be true?

	∆Gº=0
	∆Gº=1
	∆Gº>0
	∆Gº<0

2. What is the value of Keq if ∆Gº=-9.48 kJ (R=8.314 J/mol-K)?

	6700
	1.00
	1.01
	45.9

3. Calculate ∆Gº for the reaction, C₃H₈(g)+5O₂(g)→3CO₂(g)+4H₂O(g), where ∆G_fº=-23.5 kJ/mol for CH₄(g), -394 kJ/mol for CO₂(g), and -229 kJ/mol for H₂O(g):

	600 kJ
	-600 kJ
	-2070 kJ
	2070 kJ

Answer 1

1.

Keq= 8.63x10^-4

G0 = - RT lnKeq

Keq is a positive value

G0 getting the negative value

So, G0 is less than zero

G0<0

The answer is D

2)

G0 = - 9.48 KJ

R= 8.314 J/mol-k

T = 25C = 25+273=298K

G0 = - RT lnKeq

- 9.48x10^3 = - 8.314x273xlnKeq

lnKeq = 3.826

Keq= e^3.826

Keq= 45.88

Keq= 45.9

3)

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g

G0rxn = [ 3xG0f of CO2 + 4xG0f of H2O] - [ G0f of C3H8 + 5xG0f of O2]

G0rxn =[ 3 x( - 394) + 4x( -229)] - [ - 23.5 + 5 x 0.0]

G0rxn = - 2074.5 KJ

according to the given values the answer is

G0rxn = - 2074.5 KJ

(OR)

The answer is C