Question

The$\mathit{\Delta }{\mathit{G}}_{\mathbf{f}}^{\mathbf{°}}$of${\mathbf{\text{Si}}}_{\mathbf{3}}\mathbf{}{\mathbf{\text{N}}}_{\mathbf{4}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$is$\mathbf{-}\mathbf{642}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{\text{kJ}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$. Use this fact and the data in Appendix D to computelocalid="1663657851606" $\mathbf{∆}\mathit{G}\mathbf{°}$of the reaction.

$\mathbf{3}{\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Si}}}_{\mathbf{3}}\mathbf{}{\mathbf{\text{N}}}_{\mathbf{4}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathbf{3}{\mathbf{\text{SiO}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{\text{quartz}}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{}{\mathbf{\text{N}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{3}\mathbf{\text{C}}\mathbf{(}\mathit{s}\mathbf{,}\mathbf{\text{gr}}\mathbf{)}$. localid="1663658353555" $\mathbf{25}\mathbf{°}\mathit{C}$

Short answer

The free energy change the reaction$\Delta G°=-744.33\text{kJ}$ .

Reactions with a negative$\Delta G°$ release energy, which means that they can proceed without an energy input (are spontaneous).

Step-by-step solution

Given data

The given chemical equation is:.

$3{\text{CO}}_2\left(g\right)+{\text{Si}}_3{\text{N}}_4\left(s\right)\to 3{\text{SiO}}_2\left(\text{quartz}\right)+2{\text{N}}_2\left(g\right)+3\text{C}(s,\text{gr})$

$\Delta G_f^{°}$of${\text{Si}}_3{\text{N}}_4\left(s\right)$ is$-642.6\text{kJ}{\text{mol}}^{-1}$ .

$\Delta G_f^{°}\left[{\text{SiO}}_2(\right.$quartz$\left.)\right]=-856.67{\text{kJmol}}^{-1}$

$\Delta G_f^{°}\left[{\text{N}}_2\left(g\right)\right]=0{\text{kJmol}}^{-1}$

$\Delta G_f^{°}\left[\text{C}\right(s,gr\left)\right]=0{\text{kJmol}}^{-1}$

$$\begin{gathered} \Delta G_f^{°}\left[{\text{CO}}_2\left(g\right)\right]=-394.36{\text{kJmol}}^{-1} \\ \Delta G_f^{°}\left[{\text{Si}}_3{\text{N}}_4\left(s\right)\right]=-642.6{\text{kJmol}}^{-1} \end{gathered}$$

$∆G_f^{°}$represents the Standard Gibbs free energy of formation.

Concept used

The energy associated with a chemical reaction that can be used to do work. The free energy of a system is the sum of its enthalpy plus the temperature's product and the system's entropy.

Reactions with a negative$\mathbf{∆}\mathit{G}\mathbf{°}$release energy, which means that they can proceed without an energy input (are spontaneous). In contrast, reactions with a positive$\mathbf{∆}\mathit{G}\mathbf{°}$need an input of energy in order to take place (are non-spontaneous).

Calculate the value of ∆G°

The $∆G°$has been calculated as follows:

$\Delta G°={\left(\Delta G_f^{°}\right)}_{\text{products}}-{\left(\Delta G_f^{°}\right)}_{\text{reactants}}$ ......... (1)

The given chemical equation is:

$3{\text{CO}}_2\left(g\right)+{\text{Si}}_3{\text{N}}_4\left(s\right)\to 3{\text{SiO}}_2\left(\text{quartz}\right)+2{\text{N}}_2\left(g\right)+3\text{C}(s,\text{gr})$

$\Delta G_f^{°}\left[{\text{SiO}}_2(\right.$quartz$\left.)\right]=-856.67{\text{kJmol}}^{-1}$

$\Delta G_f^{°}\left[{\text{N}}_2\left(g\right)\right]=0{\text{kJmol}}^{-1}$

$\Delta G_f^{°}\left[\text{C}\right(s,gr\left)\right]=0{\text{kJmol}}^{-1}$

$$\begin{gathered} \Delta G_f^{°}\left[{\text{CO}}_2\left(g\right)\right]=-394.36{\text{kJmol}}^{-1} \\ \Delta G_f^{°}\left[{\text{Si}}_3{\text{N}}_4\left(s\right)\right]=-642.6{\text{kJmol}}^{-1} \end{gathered}$$

$∆G_f^{°}$represents the Standard Gibbs free energy of formation.

Substitute these values in equation(1)

role="math" localid="1663660019256" $$\begin{gathered} \Delta G°=3\Delta G_f^{°}\left[{\text{SiO}}_2(\right.quartz)\left.)\right]+2\Delta G_f^{°}\left[{\text{N}}_2\left(g\right)\right]+3\Delta G_f^{°}[\text{C}(s,gr)]-3\Delta G_f^{°}\left[{\text{CO}}_2\left(g\right)\right]+\Delta G_f^{°}\left[{\text{Si}}_3{\text{N}}_4\left(s\right)\right] \\ =3\text{mol}\times \left(-856.67{\text{kJmol}}^{-1}\right)+2\times 0+3\times 0-3\text{mol}\times \left(-394.36{\text{kJmol}}^{-1}\right)-1\text{mol}\times \left(-642.6{\text{kJmol}}^{-1}\right) \\ =-2570.01\text{kJ}+1183.08\text{kJ}+642.6\text{kJ} \\ =-744.33\text{kJ} \end{gathered}$$