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?

1 Answer
Andy Wolff ยท Ernest Z.
Nov 18, 2016

-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

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