Answer the questions by looking at the following reaction between nitrogen oxides.
N2O(g) + NO2(g) → 3NO(g)
(a) Find the values of ΔH°, ΔS°, ΔG°
(b) Assuming that ΔH° and ΔS° do not change with temperature, calculate ΔG° at 800 K. 800 K, is this reaction spontaneous under standard conditions?
(c) Calculate ΔG° at 1000 K. Is this reaction spontaneous at this temperature and at standard conditions?
(a) To find the values of ΔH°, ΔS°, and ΔG° for the reaction, we'll need the standard enthalpy change (ΔH°) and standard entropy change (ΔS°) for each species involved.
The standard enthalpy change for the reaction can be obtained by using the standard enthalpy of formation values for each compound involved.
Standard enthalpy change (ΔH°):
N2O(g): ΔH°f = 81.6 kJ/mol
NO2(g): ΔH°f = 33.2 kJ/mol
NO(g): ΔH°f = 90.3 kJ/mol
ΔH° = ΣΔH°(products) - ΣΔH°(reactants)
= (3 × ΔH°f(NO)) - (ΔH°f(N2O) + ΔH°f(NO2))
= (3 × 90.3 kJ/mol) - (81.6 kJ/mol + 33.2 kJ/mol)
= 270.9 kJ/mol - 114.8 kJ/mol
= 156.1 kJ/mol
The standard entropy change (ΔS°) can be found by subtracting the sum of the entropies of the reactants from the sum of the entropies of the products.
Standard entropy change (ΔS°):
N2O(g): ΔS° = 219.6 J/(mol·K)
NO2(g): ΔS° = 240.0 J/(mol·K)
NO(g): ΔS° = 210.6 J/(mol·K)
ΔS° = ΣΔS°(products) - ΣΔS°(reactants)
= (3 × ΔS°(NO)) - (ΔS°(N2O) + ΔS°(NO2))
= (3 × 210.6 J/(mol·K)) - (219.6 J/(mol·K) + 240.0 J/(mol·K))
= 631.8 J/(mol·K) - 459.6 J/(mol·K)
= 172.2 J/(mol·K)
The standard Gibbs free energy change (ΔG°) can be calculated using the formula:
ΔG° = ΔH° - TΔS°
Substituting the known values:
ΔG° = 156.1 kJ/mol - (800 K × 0.1722 kJ/(mol·K))
= 156.1 kJ/mol - 137.76 kJ/mol
= 18.34 kJ/mol
(b) To calculate ΔG° at 800 K, we can use the equation from part (a) and substitute the given temperature:
ΔG° = 156.1 kJ/mol - (800 K × 0.1722 kJ/(mol·K))
= 156.1 kJ/mol - 137.76 kJ/mol
= 18.34 kJ/mol
The reaction is spontaneous at 800 K under standard conditions if ΔG° is negative. In this case, ΔG° is positive (18.34 kJ/mol), so the reaction is not spontaneous at 800 K under standard conditions.
(c) To calculate ΔG° at 1000 K, we can again use the equation from part (a) and substitute the given temperature:
ΔG° = 156.1 kJ/mol - (1000 K × 0.1722 kJ/(mol·K))
= 156.1 kJ/mol - 172.2 kJ/mol
= -16.1 kJ/mol
The reaction is spontaneous at 1000 K under standard conditions if ΔG° is negative. In this case, ΔG° is negative (-16.1 kJ/mol), so the reaction is spontaneous at 1000 K under standard conditions.
To find the values of ΔH°, ΔS°, and ΔG° for the given reaction, we can use the standard enthalpy of formation (ΔH°f) and standard entropy (ΔS°) values for the compounds involved. The standard Gibbs free energy change (ΔG°) can then be calculated using the equation:
ΔG° = ΔH° - TΔS°
where T is the temperature in Kelvin.
(a) First, we need to look up the standard enthalpy of formation and standard entropy values for N2O(g), NO2(g), and NO(g). Let's assume the following values (in kJ/mol):
ΔH°f(N2O(g)) = 82.05
ΔH°f(NO2(g)) = 33.18
ΔH°f(NO(g)) = 90.25
ΔS°(N2O(g)) = 220.79
ΔS°(NO2(g)) = 240.03
ΔS°(NO(g)) = 210.70
Using these values, we can calculate ΔH°, ΔS°, and ΔG° for the reaction:
ΔH° = (3 * ΔH°f(NO(g))) - (ΔH°f(N2O(g)) + ΔH°f(NO2(g)))
= (3 * 90.25) - (82.05 + 33.18)
= 206.77 kJ/mol
ΔS° = (3 * ΔS°(NO(g))) - (ΔS°(N2O(g)) + ΔS°(NO2(g)))
= (3 * 210.70) - (220.79 + 240.03)
= 52.17 J/(mol·K)
ΔG° = ΔH° - TΔS°
(b) To calculate ΔG° at 800 K, we substitute the values into the equation:
T = 800 K
ΔG° = 206.77 - (800 * 0.05217)
= 206.77 - 41.736
= 165.034 kJ/mol
Since ΔG° is positive (165.034 kJ/mol), the reaction is non-spontaneous under standard conditions at 800 K.
(c) To calculate ΔG° at 1000 K, we substitute the values into the equation:
T = 1000 K
ΔG° = 206.77 - (1000 * 0.05217)
= 206.77 - 52.17
= 154.60 kJ/mol
Again, since ΔG° is positive (154.60 kJ/mol), the reaction is non-spontaneous both at 1000 K and under standard conditions.
I suppose I don't understand why you're having trouble unless you're looking for someone to do all of the work.
a. Look up the values in a table. If you don't have a table look them up on the web.
b. dG = dH - TdS. Plug in the values from a. The reaction is spontaneous @ 800 K if dG is negative.
c. Change 800 K to 1000 K and recalculate b.
Post your wor if you get stuck but include the values you found in part a and your calculations for b and c..