Question

For the following reaction: ${{K}_{P}}=1.7\times {{10}^{7}}$ at 25$^{o}C$
$A{{g}^{+}}(aq)+2N{{H}_{3}}(g)\rightleftarrows {{\left[ Ag{{(N{{H}_{3}})}_{2}} \right]}^{+}}$
What value of $\Delta {{G}^{o}}$ in kJ?
(A) -41.2
(B) -17.9
(C) +17.9
(D) +41.2

Answer 1

Relation between Gibbs free energy ($\Delta {{G}^{o}}$) of a reaction and equilibrium constant (${{K}_{p}}$) is given as $\Delta {{G}^{o}}=-RT\ln ({{K}_{p}})$.
Since ${{\ln }_{e}}=2.303{{\log }_{10}}$, expression for $\Delta {{G}^{o}}$ is also written as $\Delta {{G}^{o}}=-2.303RT\log ({{K}_{p}})$.
Value of universal gas constant, R = $8.314J{{K}^{-1}}mo{{l}^{-1}}$

Complete step by step answer:
The given chemical reaction is
$A{{g}^{+}}(aq)+2N{{H}_{3}}(g)\rightleftarrows {{\left[ Ag{{(N{{H}_{3}})}_{2}} \right]}^{+}}$
We have been given the equilibrium constant of the reaction in terms of pressure, ${{K}_{P}}=1.7\times {{10}^{7}}$
Given, temperature of the reaction, T = 25$^{o}C$.
To convert temperature in Celsius ($^{o}C$) to Kelvin (K),
T (K) = T ($^{o}C$ ) + 273.15
T (K) = 25 + 273.15 = 298.15 K

We know that Gibbs free energy of a reaction is given as, $\Delta {{G}^{o}}=-2.303RT\log ({{K}_{p}})$
Substituting R = $8.314J{{K}^{-1}}mo{{l}^{-1}}$, T = 298.15 K and ${{K}_{P}}=1.7\times {{10}^{7}}$ in the above expression, we get
$\begin{aligned} & \Delta {{G}^{o}}=-2.303RT\log ({{K}_{p}}) \\\ & \Delta {{G}^{o}}=-2.303\times 8.314J{{K}^{1}}mo{{l}^{-1}}\times 298.15K\times \log (1.7\times {{10}^{7}}) \\\ & \Delta {{G}^{o}}=-5708.72\times \log (1.7\times {{10}^{7}}) \\\ \end{aligned}$
Applying $\log (mn)=\log m + \log n$ and $\log {{m}^{n}}=n\log m$, we can simplify the above equation as
$\begin{aligned} & \Delta {{G}^{o}}=-5708.72Jmo{{l}^{-1}}\times (\log 1.7+\log {{10}^{7}}) \\\ & \Delta {{G}^{o}}=-5708.72Jmo{{l}^{-1}}\times (\log 1.7+7\log 10) \\\ \end{aligned}$
On substituting ${{\log }_{10}}10=1$and log (1.7) = 0.23045, the above equation becomes
$\begin{aligned} & \Delta {{G}^{o}}=-5708.72Jmo{{l}^{-1}}\times (0.23045+7\times 1) \\\ & \Delta {{G}^{o}}=-5708.72Jmo{{l}^{-1}}\times 7.23045 \\\ & \Delta {{G}^{o}}=-41276.60Jmo{{l}^{-1}} \\\ \end{aligned}$

To convert the free energy in kilojoules, divide it by ${{10}^{3}}$.
Since 1 kJ = 1000 J or 1 J = $\dfrac{1}{1000}$ kJ.
Thus, we have -41276.60 J = $\dfrac{-41276.60}{1000}$ kJ = -41.27660 kJ.
Therefore, the $\Delta {{G}^{o}}$ of the given reaction = $41.27660kJmo{{l}^{-1}}$.
So, the correct answer is “Option A”.

Note: Note that when ${{K}_{p}}>1$, then $\Delta {{G}^{o}}<0$. Negative value of Gibbs free energy means that the reaction is spontaneous and proceeds in the forward direction. Carefully solve the question to avoid calculation errors.