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)

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)

he would use the colour black

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.