Free Energy - Advanced Higher Chemistry

Question

Given these reaction values for oxides of nitrogen:

4NO(g) → 2N2O(g) + O2(g) ΔG◦ = -139.56 kJ

2NO(g) + O2(g) → 2NO2(g) ΔG◦ = -69.70 kJ

a) Calculate ΔG◦ for this reaction: 2N2O(g) + 3O2(g) 4NO2(g)

b) Say whether the equilibrium position favours reactants or products.

Can anyone help me? I understand part b) I just don't know how to find a). The answer for part a) is apparently +0.16 kJ

Answer 1

Given these reaction values for oxides of nitrogen:

4NO(g) → 2N₂O(g) + O₂(g) ΔG◦ = -139.56 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔG◦ = -69.70 kJ

a) Calculate ΔG◦ for this reaction: 2N₂O(g) + 3O2(g) 4NO₂(g)

ΔG◦ = ΔG (Products) - ΔG (Reactants)

You must know this equation. To get Gibbs Free Energy for producs you must also know that any element in it's single state for example Fe(s) or O₂ (g) have 0 = ΔG

4NO(g) → 2N₂O(g) + O₂(g) ΔG◦ = -139.56 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔG◦ = -69.70 kJ

Let's eliminate O₂(g) because the O₂ (g) have 0 = ΔG

4NO(g) → 2N₂O(g) ΔG◦ = -139.56 kJ

2NO(g) → 2NO₂(g) ΔG◦ = -69.70 kJ

Remember in these equations the desired 2N₂O(g) and 4NO₂(g) are products. We must make them reactants by flipping the equation and switching - to +.

2N₂O(g) → 4NO(g) ΔG◦ = +139.56 kJ

2NO₂(g) → 2NO(g) ΔG◦ = +69.70 kJ

Now keep track coefficients you want 4 NO₂(g) not 2 NO₂(g). Simply multiply everything by 2 to get 4NO₂(g)

4N₂O(g) → 8NO(g) ΔG◦ = +279.12 kJ

Because this equation includes for both products and reactants ΔG◦ simply add to get desired reactants.

+279.12 kJ (4N₂O_(g)) + +69.70 kJ (2NO_2(g) ) + 0 (O_{2 (g)}) =

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Answer 2

This is called Hess's law and what you need to do is manipulate the two equations given to give you the equation found in part a.

Since 2N₂O is found as a reactant for part a that means the first equation must be flipped so it is a reactant giving

2N₂O + O₂ --> 4NO and because of this the sign changes to +139.56kj

Now you see that 4NO₂ is a product for part a. the only reaction with NO₂ is the second one but it has 2NO₂ as a product so this equation and therefore its energy term must be doubled. giving

4NO + 2O₂ --> 4NO₂ 2(-69.70kj) = -139.4kj

Now you sum up the 2 equations giving

2N₂O + 4NO + 3O₂ -->4NO + 4NO₂ Notice there is 4NO on both sides so like spectator ions they cancel out giving the equation you find in part a. Now you sum up the energy.

139.56kj + (-139.4kj) = 0.16kj.and in this case favors the reactants since the Gibb's value is positive.