Question

The equilibrium constant for a reaction is $ 10 $ . What will be the value of $ \Delta {G^ \circ } $ when $ R{\text{ = 8}}{\text{.314 J}}{{\text{K}}^{ - 1}}mo{l^{ - 1}} $ , $ T{\text{ = 300 K}} $ . $ $

Answer 1

Hint: We will use the relationship between the standard free energy change of the reaction and the equilibrium constant of the reaction. Gibbs free energy is the maximum work that can be derived from a system at constant temperature and pressure.

Complete answer:
The relation between the free energy change is given by:
 $ \Delta G{\text{ = }}\Delta {{\text{G}}^ \circ }{\text{ + RT lnQ}} $
Here,
 $ \Delta G{\text{ = }} $ Free energy change of the system
 $ \Delta {G^ \circ }{\text{ = }} $ Standard free energy change of the system
 $ {\text{R = }} $ Universal gas constant, $ {\text{ = 8}}{\text{.314 J}}{{\text{K}}^{ - 1}}mo{l^{ - 1}} $
 $ {\text{T = }} $ $ 300{\text{ K}} $
 $ Q{\text{ = }} $ Reaction Quotient
For the gases at the equilibrium the reaction quotient becomes equal to the equilibrium constant. Therefore the equilibrium will replace the term reaction quotient in the above equation. Also the value of $ \Delta G $ becomes zero. This means that at equilibrium for gases the free energy change is always equal to zero. Therefore the above equation will finally reduce to:
 $ {\text{0 = }}\Delta {{\text{G}}^ \circ }{\text{ + RT lnK}} $
 $ \Delta {{\text{G}}^ \circ }{\text{ = - RT lnK}} $
We can convert the $ \ln K $ into $ \log K $ by multiplying the above equation by $ 2.303 $ . Hence on multiplying it with $ 2.303 $ we get the finalised equation as:
 $ \Delta {{\text{G}}^ \circ }{\text{ = - 2}}{\text{.303 RT log K}} $
Here we are given an equilibrium constant of $ 10 $ . Therefore on putting the values we get,
 $ \Delta {{\text{G}}^ \circ }{\text{ = - 2}}{\text{.303 }} \times {\text{ 8}}{\text{.314 }} \times {\text{ 300 log 10}} $
We know that $ \log 10{\text{ = 1}} $ , therefore
 $ \Delta {{\text{G}}^ \circ }{\text{ = - 2}}{\text{.303 }} \times {\text{ 8}}{\text{.314 }} \times {\text{ 300 }} $
 $ \Delta {{\text{G}}^ \circ }{\text{ = - 5744 Jmo}}{{\text{l}}^{ - 1}} $
It can be written in kilo-joule. Therefore the standard change in free energy is:
 $ \Delta {{\text{G}}^ \circ }{\text{ = - 5}}{\text{.744 kJmo}}{{\text{l}}^{ - 1}} $
The negative change in free energy means that the reactants have more free energy than the products.

Note:
We should convert the $ \ln K $ into $ \log K $ to make our calculations easy. The units of free energy are the same as that of energy. The only difference is that free energy is energy per mole. This free energy can be both positive and negative, so do not change the sign. A positive change in energy means the products have more free energy than the reactants.