Question

Consider the production of ammonia from nitrogen and hydrogen,

$N_2+3H_2\to 2N{H}_3$

at 298 K and 1 bar. From the values of ΔH and S tabulated at the back of this book, compute ΔG for this reaction and check that it is consistent with the value given in the table.

Short answer

The value of ΔG = -32972.5J

Step-by-step solution

Given Information

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

Explanation

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

ΔG = ΔH - TΔS ........................................(1)

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

$\Delta H=2\Delta H_{{\mathrm{NH}}_3}-\Delta H_{{\mathrm{N}}_2}-3\Delta H_{{\mathrm{H}}_2}$

Now substitute the value from the table , we get

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

Change in entropy for the reaction

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

Substitute values

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

Substitute calculated values in equation (1), we get
$$\begin{gathered} \Delta G=\left(-92.2\times {10}^3\mathrm{J}\right)-(298\mathrm{K})\left(-198.75{\mathrm{JK}}^{-1}\right) \\ =-32972.5\mathrm{J} \end{gathered}$$