When K is measured for a reaction, the values obtained are 3.1 at 25°C and 7.42 at 35°C. Using the equation lnK=-(∆H°/R)(1/T)+(∆S°/R) (in the form y=mx+b) calculate ∆H° and ∆S° for the reaction PbCl2(s) --> Pb2+(aq) + Cl-(aq)

Question

When K is measured for a reaction, the values obtained are 3.1 at 25°C and 7.42 at 35°C. Using the equation lnK=-(∆H°/R)(1/T)+(∆S°/R) (in the form y=mx+b) calculate ∆H° and ∆S° for the reaction PbCl2(s) --> Pb2+(aq) + Cl-(aq)

Answer 1

b = (∆S°/R)

so
ln 3.1 = m (1/298) + b
ln 7.42 = m(1/308) + b
----------------------------subtract
ln (3.1/7.42) = m (1/298 - 1/308)
-.873 = .000109 m
m = - 8009 = - (∆H°/R)
so
∆H° = 8009 R
go back and get ∆S°