Question

At $27^\circ C$, \[{\text{4N}}{{\text{H}}_{\text{3}}}\left( {\text{g}} \right){\text{ + 5}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ }} \to {\text{ 4NO}}\left( {\text{g}} \right){\text{ + 6}}{{\text{H}}_{{\text{2}}}}{\text{O}}\left( {\text{l}} \right)\]
Given, \[\Delta H = 9080Jmo{l^{ - 1}}\] and \[\Delta S = 35.7J{K^{ - 1}}mo{l^{ - 1}}\].
The given reaction is ________ (spontaneous/ non-spontaneous).

Answer 1

Hint: We need to remember that the Gibbs free energy is a thermodynamic term which tells us about the maximum reversible work that can be performed by any thermodynamic system at constant temperature and pressure. If the Gibbs free energy is positive then the process is nonspontaneous and if it is negative then the process is spontaneous.
Formula used:
$\Delta {\text{G = }}\Delta {\text{H - T}}\Delta {\text{S}}$
Where,
$\Delta G$ - Change in Gibbs free energy
$\Delta H$ - Change in enthalpy
$T$ - Temperature of system (In kelvin)
$\Delta S$ - Change in enthalpy

Complete step by step answer:
This question is related to Gibbs free energy, so let’s discuss Gibbs free energy. Gibbs free energy can also be defined as the maximum amount of non-expansion work that can be extracted from a thermodynamically closed system. It is given by the formula $\Delta {\text{G = }}\Delta {\text{H - T}}\Delta {\text{S}}$
So, if temperature and $\Delta {\text{S}}$is higher then automatically the Gibbs energy will be low. We are given with \[\Delta H = 9080Jmo{l^{ - 1}}\] and \[\Delta S = 35.7J{K^{ - 1}}mo{l^{ - 1}}\] and \[T = {27^o}C = 300K\] . Putting these values in the equation we get,
\[\Delta G = 9080Jmo{l^{ - 1}}-300 \times \left( {35.7} \right)Jmo{l^{ - 1}}\]
On simplification we get,
\[\;\Delta G = - 1630Jmo{l^{ - 1}}\]
From the calculation value the Gibb’s free energy is less than zero.
\[\Delta G < 0\],
$\therefore $Hence the reaction/process is spontaneous in nature.

Note:
We can calculate the Gibbs free energy by using the given formula or it can be calculated by graphical method. This energy tells us whether the reaction or process that is going to happen will be spontaneous or nonspontaneous. It also helps in calculating or deriving the Nernst equation which is also an important thermodynamic equation. We need to remember that the Gibbs free energy is not a path function which follows state function.