Question

Gold ions form complexes with cyanide ion according to the equation: `"Au"^(+)(aq) + 2 "CN"^(-) -> ["Au"("CN")_2]^(-)`; `K_f = 2xx10^38` What is the standard state free energy for this reaction?

Answer 1

-218. kJ

Explanation:

For this you need to know the relationship between the free energy of the reaction (also called the Gibbs Free Energy, `Delta G`) and the equilibrium constant, `K_(f)`:

`Delta G = -RT*lnK_(f)`

You already have `K_f`, so all we need is the gas constant, R (8.314 J/(mol*K), and the temperature.

For thermodynamics, the standard state is always `25^o C` and 1 atm pressure. So T (and remember that all thermodynamic calculations use absolute temperature or kelvins in the metric and SI system) is therefore degrees C + 273.15 K or 298.15 K.

So now all we do is plug in the numbers:

`Delta G = -(8.314 J/(mol*K) * 298.15 * ln( 2 xx 10^(38)))`

`Delta G = -218610.47897 J`

From the above we see that the `ln(2 xx 10^(38))` will limit our significant figures to 3. So...

`Delta G = -218 kJ`