Question

Given the following data:

$$\begin{gathered} \mathbf{2}{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{6}\mathbf{\left(}\mathbf{l}\mathbf{\right)}}\mathbf{+}\mathbf{15}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{12}\mathit{C}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{\left(}\mathbf{l}\mathbf{\right)}} \\ {\mathit{C}}_{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathit{C}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}} \\ {\mathit{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{O}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{\left(}\mathbf{l}\mathbf{\right)}} \end{gathered}$$

Calculate $\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}$ for the reaction $\mathbf{6}{\mathit{C}}_{\mathbf{\left(}\mathbf{s}\mathbf{\right)}}\mathbf{+}\mathbf{3}{\mathit{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }{\mathit{C}}_{\mathbf{6}}{\mathit{H}}_{\mathbf{6}\mathbf{\left(}\mathbf{l}\mathbf{\right)}}$

$$\begin{gathered} \Delta G^o=-6399kJ \\ \Delta G^o=-394kJ \\ \Delta G^o-237kJ\mathrm{d} \end{gathered}$$

Short answer

$124.5kJ/\text{mol}$is the value of $\Delta G^o$

Step-by-step solution

Multiply the reactions

Multiply the second equation by 6 in the reactant side.

$$\begin{gathered} 6\times C_{\left(s\right)}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)} \\ 6C_{\left(s\right)}+6O_{2\left(g\right)}\to 6C{O}_{2\left(g\right)} \end{gathered}$$

Consider this equation one.

Multiply the third equation by $3H_2$ in the reactant side

$$\begin{gathered} 3\times H_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to H_2{O}_{\left(l\right)} \\ 3H_{2\left(g\right)}+\frac{3}{2}{O}_{2\left(g\right)}\to 3H_2{O}_{\left(l\right)} \end{gathered}$$

Consider this equation two.

Multiply the first equation by 1/2 to gain one $C_6{H}_{6\left(l\right)}$ on the product side and reverse the reaction.

$$\begin{gathered} \frac{1}{2}\times (12C{O}_{2\left(g\right)}+6H_2{O}_{\left(l\right)})\to 2C_6{H}_{6\left(l\right)}+15O_{2\left(g\right)} \\ 6C{O}_{2\left(g\right)}+36H_2{O}_{\left(l\right)}\to C_6{H}_{6\left(l\right)}+\frac{15}{2}{O}_{2\left(g\right)} \end{gathered}$$

Adding the reactions

Now, add the equations

$$\begin{gathered} \frac{6C_{\left(s\right)}+6O_{2\left(g\right)}\to 6C{O}_{2\left(g\right)} \\ 3H_{2\left(g\right)}+\frac{3}{2}{O}_{2\left(g\right)}\to 3H_2{O}_{\left(l\right)} \\ 6C{O}_{2\left(g\right)}+36H_2{O}_{\left(l\right)}\to C_6{H}_{6\left(l\right)}+\frac{15}{2}{O}_{2\left(g\right)}}{6C_{\left(s\right)}+3H_{2\left(g\right)}\to C_6{H}_{6\left(l\right)}} \end{gathered}$$

Then, apply the condition.

$$\begin{gathered} \Delta G^o=\left[6(-394)+3(-237)-\frac{1}{2}(-6399)\right] \\ \Delta G^o=(-2364-711+3199.5) \\ \Delta G^o=124.5kJ/mol\mathrm{d} \end{gathered}$$

Conclusion

Therefore, $124.5kJ/\text{mol}$ is the value of $∆{\mathrm{G}}^{°}$