Title: Practice Questions Section 2
Author: Peggy Lawson

Unit 1: Thermodynamics	Chemistry 30

Practice Questions Section 3.3
Gibbs Free Energy

1. 	Calculate ΔG at 25°C for the following reaction, by first calculating ΔH and ΔS. Once you've found ΔH and ΔS, solve for ΔG using the formula:

		ΔG = ΔH - T ΔS

	Also - will this reaction be spontaneous at this temperature?

		CH_3CO_2H _(l) + 2 O_2 _(g) → 2 CO_2 _(g) + 2 H_2O _(g)

2.	Again find ΔG at 25°C for the reaction

		CH_3CO_2H _(l) + 2 O_2 _(g) → 2 CO_2 _(g) + 2 H_2O _(g)

	This time using the Table of Thermochemical Data and the formula:  ΔG = ΣΔG° _products - ΣΔG° _reactants 

3.	For the reaction Fe_2O_3 _(s) + 3 CO _(g) → 2 Fe _(s) + 3 CO_2 _(s) 

ΔG° = -31.3 kJ. Calculate the standard free energy of formation of the ferric oxide, Fe_2O_3,
 if ΔG°_f of CO = -137 kJ/mol and ΔG°_f of CO_2 = -394 kJ/mol.

Practice Questions Section 3.3
Gibbs Free Energy	Answers

1. 	Calculate ΔG at 25°C for the following reaction, by first calculating ΔH and ΔS. Once you've found ΔH and ΔS, solve for ΔG using the formula:

		ΔG = ΔH - T ΔS

	Also - will this reaction be spontaneous at this temperature?

		CH_3CO_2H _(l) + 2 O_2 _(g) → 2 CO_2 _(g) + 2 H_2O _(g)

	Solution:

		Step 1 - Calculate ΔH

CH_3CO_2H _(l) | + | 2 O_2 _(g) | → | 2 CO_2 _(g) | + | 2 H_2O _(g)
-484.5 | + | 2 × 0 | | 2 × (-393.5) | + | 2 × (-241.8)
-484.5 | | -1270.6

			ΔH 	= ΣΔH° _products - ΣΔH° _reactants 

				=   -1270.6 - (-484.5)
	
				=   -786.1 kJ	Answer

		Step 2- Calculate ΔS 

CH_3CO_2H _(l) | + | 2 O_2 _(g) | → | 2 CO_2 _(g) | + | 2 H_2O _(g)
159.8 | + | 2 × 205.1 | | 2 × 213.7 | + | 2 × 188.8
570.0 | | 805.0

			ΔS	= ΣΔS° _products - ΣΔS° _reactants 

				= 805.0 - (570.0)

				= 235.0 J/K = 0.235 kJ/K		Convert to kJ / K for calculating ΔG

	Step 3 - Calculate ΔG Be sure to convert 25°C into K and ΔS into kJ/K

			K	= C + 273
				= 25 + 273
				= 298 K

			ΔG	= ΔH - T ΔS

				= -786.1 – (298.0 × 0.235)
	
				= -856.1 kJ	answer

		Because ΔG is negative, the reaction is spontaneous at this temperature.

2.	Again find ΔG at 25°C for the reaction

		CH_3CO_2H _(l) + 2 O_2 _(g) → 2 CO_2 _(g) + 2 H_2O _(g)

	This time using the Table of Thermochemical Data and the formula:  ΔG = ΣΔG° _products - ΣΔG° _reactants 

	Solution:

		Look up ΔG values for all reaction participants. Multiply by coefficients from the balanced equation. Find totals for the reactant and product sides of the equation: 

CH_3CO_2H _(l) | + | 2 O_2 _(g) | → | 2 CO_2 _(g) | + | 2 H_2O _(g)
-389.9 | + | 2 × 0 | | 2 × (-394.4) | + | 2 × (-228.6)
-389.9 | | -1246.0

		ΔG	= ΣΔG° _products - ΣΔG° _reactants 
	
			= -1246.0 - (-389.9)

		ΔG = -856.1 kJ	answer

3.	For the reaction Fe_2O_3 _(s) + 3 CO _(g) → 2 Fe _(s) + 3 CO_2 _(s) 

ΔG° = -31.3 kJ. Calculate the standard free energy of formation of the ferric oxide, Fe_2O_3,
 if ΔG°_f of CO = -137 kJ/mol and ΔG°_f of CO_2 = -394 kJ/mol.

	Solution:

	This time we are given the value ΔG°for the entire reaction, and need to find ΔG°_f for one of the reaction participants, Fe_2O_3. Let's let that unknown equal x:  (Yes, you could look up the answer in the Table of Thermochemical Data, but let’s use that to check our answer at the end)
		
Fe_2O_3 | + | 3 CO | → | 2 Fe | + | 3 CO_2
x | + | 3 × (-137) | | 2 × 0 | + | 3 × (-394)
-411 | | -1182
	
	Next, set up our formula for ΔG° and substitute in the values we know, then solve for x: 

ΔG = ΣΔG° _products - ΣΔG° _reactants 

		-31.3 = (-1182) - (x - 411)

		-31.3 = -1182 - x + 411

		-31.3 = -771- x

		x = -771 + 31.3 =  -740

Answer - ΔG°_f for Fe_2O_3 = -740 kJ/mol