# 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?

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, ${K}_{f}$:

$\Delta G = - R T \cdot \ln {K}_{f}$

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:

$\Delta G = - \left(8.314 \frac{J}{m o l \cdot K} \cdot 298.15 \cdot \ln \left(2 \times {10}^{38}\right)\right)$

$\Delta G = - 218610.47897 J$

From the above we see that the $\ln \left(2 \times {10}^{38}\right)$ will limit our significant figures to 3. So...

$\Delta G = - 218 k J$