Question

A chemist fills a reaction vessel with 0.247 M lead(II) (Pb2+) aqueous solution, 0.758 M bromide (Br-) aqueous solution, and 0.109 g lead (II) bromide (PbBr2) solid at a temperature of 20.0 oC. Under these conditions, calculate the reaction free energy deltaG for the following chemical reaction:
Pb2+(aq) + 2Br-(aq) <----> PbBr2(s)

Answer 1

Answer:

ΔG = -24.7kJ/mol

Explanation:

ΔG° of

Pb2+(aq) + 2Br-(aq) ⇄ PbBr2(s)

is:

ΔG° PbBr2 - (2*ΔG°Br- + ΔG°Pb2+)

-261.9kJ/mol - (2*-104.0kJ/mol + -24.4kJ/mol) =  

-29.5kJ/mol

ΔG of the reaction is:

ΔG = ΔG° + RT lnQ

Where R is gas constant (8.314x10⁻³kJ/molK)

T is absolute temperature (20°C + 273.15 = 293.15K)

Q is reaction quotient = 1 / [Pb²⁺][Br⁻]²

Replacing:

ΔG = -29.5kJ/mol + 8.314x10⁻³kJ/molK*293.15K ln(1 / [Pb²⁺][Br⁻]²)

ΔG = -29.5kJ/mol + 8.314x10⁻³kJ/molK*293.15K ln(1 / [0.247M][0.758M]²)

ΔG = -24.7kJ/mol