EMERGENCY HELP PLEASE

Question

23. Calculate the equilibrium constant for the following reaction at 655 K.

           HCN(g) + 2 H₂(g) → CH₃NH₂(g)             ΔH° = -158 kJ; ΔS°= -219.9 J/K

a) 39.9            b) 13.0            c) 25.1            d) 32.6

Answer 1

There are two equations to think about here:

(1) ΔG° = ΔH° - TΔS°

(2) ΔG° = -RTlnK

Setting (1) and (2) equal to each other because ΔG° will be the same, we get:

ΔH°-TΔS° = -RTlnK

divide each side by -RT to get:

-(ΔH°-TΔS°)/RT = lnK

Then, raise each side as a power of e to get...

e^(-(ΔH°-TΔS°)/RT) = K

Now you have an equation that you can plug numbers into and solve for your equilibrium constant, K. Make sure to have the same units for everything (here I will convert enthalpy into Joules to make sure the units match with the entropy)

e^(-(-158,000J -(655K)((-219.9J*K^-1))/((8.314J*mol^-1K^-1)(655K))) = K

∴ e^2.56 = K

∴12.99 = K

With three significant figures, the answer is 13.0