Given the following information:

Question

Given the following information:

F^1- (aq): ΔG_f° = -276.48 kJ/mol

Na¹⁺ (aq): ΔG_f° = -261.87 kJ/mol

2 Na _(s) + F_2 (g) → 2 NaF _(aq)

a) Find ΔG° for this reaction using the above numbers.

b) Write the expression for Q_eq. (Note that NaF is a strong electrolyte!)

c) Find ΔG at 325.0 K for 0.01500 atm F₂ and 3.065 M NaF.

Answer 1

a) To find ΔG° which is change in Gibbs Free Energy at standard conditions

Standard Conditions:

Solutions at 1M

Gases at 1atm

Temperature at 298K

Solids/Liquids in their standard states

Use Formula

ΔG° = ∑nΔGf° (products) - ∑nΔGf°(reactants)

Elements in standard state such as Na⁺(s) and F^-(g) have a Gibbs Energy of Formation of 0kJ/mol.

∑nΔGf° (products) = 2[(-261.87 kJ/mol) + (-276.48 kJ/mol)] where 2 represents the stoichiometric coefficient

a) ΔG° = -1076.70kJ

Q_eq is the ratio of concentration of products/reactants at any point during a reaction while K_eq is the ratio of concentration of products/reactants when reaction is at equilibrium.

2 Na _(s) + F_2 (g) → 2 NaF _(aq)

NaF is a strong electrolyte meaning it dissolves completely into Na⁺ and F^- ions in solution. Therefore reaction becomes 2Na(s) + F₂(g) → 2Na⁺(aq) + 2F^-(aq). Set this up in Q_eq...

Solid Sodium Na(s) does not appear in denominator of Q_eq because their activity is constant and its concentration doesn't necessarily change during reaction. The "effective concentration" of the solid phase stays the same so we label it as 1. Since F_2(g) is a gas we express it as a pressure instead of concentration which is acceptable in a reaction quotient (Q).

b) Q_eq = [Na⁺]²[F^-]² / P_F2

So as discussed above ΔG° is change in Gibbs Energy at standard state conditions while ΔG is change in Gibbs Energy under actual current reaction conditions.

Use formula ΔG = ΔG° + RTln(Q)

T = 325K

0.01500 atm F₂ and 3.065 M NaF

Q_eq = [3.065]²[3.065]² / 0.01500 = 5886.6 (unitless)

c) ΔG = ΔG° + RTln(Q) = -1076.70kJ/mol + [0.008314(kJ/mol*K) (325K) ln(5886.6)] = -1053kJ/mol