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 · 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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