91Ó°ÊÓ

Temp, T = 298 K and the
pressure P= 1 bar.
The table from the book

Gibbs energy can be calculated by the equation below.

G=H-T S
Where, G= Gibbs energy, H= enthalpy, T= absolute temperature and S= entropy.

Lets assume there is an infinitesimal change is Gibbs energy, then

$\mathrm{Δ}G=\mathrm{Δ}H-T\mathrm{Δ}S…..........\left(1\right)$

Similarly equation for the change in enthalpy for the given reaction is written as

$\mathrm{Δ}H=2\mathrm{Δ}{H}_{{\mathrm{NH}}_3}-\mathrm{Δ}{H}_{{\mathrm{N}}_2}-3\mathrm{Δ}{H}_{{\mathrm{H}}_2}$

Now substitute the value from the table , we get

$$\begin{gathered} \mathrm{Δ}H=2(-46.11\mathrm{kJ})-0-0 \\ =-92.2\mathrm{kJ} \\ =-92.2\mathrm{kJ}\left(\frac{1000\mathrm{J}}{1\mathrm{kJ}}\right) \\ =-92.2×{10}^3\mathrm{J} \end{gathered}$$

Change in entropy for the reaction

$\mathrm{Δ}S=2\mathrm{Δ}{S}_{{\mathrm{NH}}_3}-\mathrm{Δ}{S}_{{\mathrm{N}}_2}-3\mathrm{Δ}{S}_{{\mathrm{H}}_2}$

Put the values from table, we get

$$\begin{gathered} \mathrm{Δ}S=2\left(192.45{\mathrm{JK}}^{-1}\right)-\left(191.61{\mathrm{JK}}^{-1}\right)-3\left(130.68{\mathrm{JK}}^{-1}\right) \\ =-198.75{\mathrm{JK}}^{-1} \end{gathered}$$

Now, substitute $\mathrm{Δ}S=-198.75{\mathrm{JK}}^{-1}and\mathrm{Δ}H=-92.2×{10}^3\mathrm{J}$ in the equation (1), we get$$\begin{gathered} \mathrm{Δ}G=\left(-92.2×{10}^3\mathrm{J}\right)-(298\mathrm{K})\left(-198.75{\mathrm{JK}}^{-1}\right) \\ =-32972.5\mathrm{J} \end{gathered}$$