Question

The equilibrium constant at ${25^0}C$ for the process:
$C{o^{ + 3}}_{\left( {aq} \right)} + 6N{H_{3\left( {aq} \right)}} \rightleftharpoons {\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}_{\left( {aq} \right)}$ is $2 \times {10^7}$
Calculate the value of $\Delta {G^0}$ at ${25^0}C\left( {R = 8.314J{K^{ - 1}}.mo{l^{ - 1}}} \right)$

Answer 1

The Gibbs free energy is represented by G, the change in Gibb’s free energy is represented by $\Delta G$.
The standard Gibbs free energy and Gibbs free energy are related to equilibrium constant.
The Gibbs free energy at equilibrium will be equal to zero.
Formula used:
$\Delta G = \Delta {G^0} + RT\ln K - - - - \left( 1 \right)$
Where, $\Delta G$ is change in Gibbs energy,
$\Delta {G^0}$ is change in Gibbs energy at standard conditions,
K is equilibrium constant,
R is ideal gas constant and
T is absolute temperature.

Complete answer:
Given reaction is taking place at equilibrium, it means the term change in Gibbs free energy will be Zero.
$\Rightarrow \Delta G = 0$
Equilibrium reaction is a reaction in which there is no change in concentration of products and reactions with time.
The equilibrium reaction can be represented by $\rightleftharpoons$. The two half arrows are with same length.
If one of the arrows is more lengthy than the other, then the equilibrium shifts towards that side.
Now, the relation between the standard Gibbs energy and Gibbs energy and equilibrium constant can be written as
$\Delta {G^0} = - RT\ln K - - - - \left( 2 \right)$
Given that, equilibrium constant for the process is $2 \times {10^7}$.
Ideal gas constant $\left( R \right) = 8.314J{K^{ - 1}}{mol^{ - 1}}$,
Temperature $\left( T \right) = 25 + 273 = 298K$
Substitute these values in the equation (2)
$\Delta {G^0} = - \left( {8.314} \right) \times 298 \times \ln \left( {2 \times {{10}^7}} \right)$
Thus, $\Delta {G^0} = - 41651.1Jmo{l^{ - 1}}$
But, $1kJ = 1000J$
By substituting the value of 1kJ in the above equation, we will get
$\Delta {G^0} = - 41.65kJmo{l^{ - 1}}$
Thus, the standard change in Gibbs free energy at standard conditions is $- 41.65kJmo{l^{ - 1}}$ .

Note:
While calculating the standard change in Gibbs free energy at standard conditions, the units must be considered. The Gibbs free energy is also known as Gibbs energy.
The temperature must be in Kelvins only.
The ideal gas constant must be in $J{K^{ - 1}}mo{l^{ - 1}}$.