91影视

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to measure the greatest amount of work done in a thermodynamic system when temperature and pressure remain constant.

(a)

The balanced equation is 鈥�

${\text{Cl}}_{\text{2}}{\text{(g) + Br}}_{\text{2}}\text{(g)}鈫�\text{2BrCl(g)}$

The values given are 鈥�

$$\begin{gathered} 鈭�H_{rxn}^{掳}=-\text{1.35 kJ/mol} \\ 鈭�G_f^{掳}=-\text{0.88 kJ/mol} \end{gathered}$$

Manipulate the reaction to solve for $鈭�H_f^{掳}$of $\text{BrCl(g)}$ 鈥�

$$\begin{gathered} 鈭�H_{rxn}^{掳}=\underset{}{\overset{}{鈭�}}{n}_p鈭�H_f^{掳}\left(product\right)-\underset{}{\overset{}{鈭�}}{n}_r鈭�H_f^{掳}\left(reac\tan t\right) \\ 鈭�H_{rxn}^{掳}=\left[n鈭�H_f^{掳}\left(\text{BrCl}\left(g\right)\right)\right]-\left[n鈭�H_f^{掳}\left(C{l}_2\left(g\right)\right)+n鈭�H_f^{掳}\left(B{r}_{2\left(g\right)}\right)\right] \\ n鈭�H_f^{掳}\left(B{r}_2\left(g\right)\right)=\left[n鈭�H_f^{掳}\left(C{l}_2\left(g\right)\right)+n鈭�H_f^{掳}\left(B{r}_2\left(g\right)\right)\right]+鈭�H_{rxn}^{掳} \\ 2鈭�H_f^{掳}\left(BrCl\left(g\right)\right)=\left[1\left(0\right)+1\left(30.91\right)\right]kJ/mol+\left(1.35kJ/mol\right) \\ 鈭�H_f^{掳}\left(BrCl\left(g\right)\right)=\frac{29.56kJ/mol}{2} \\ 鈭�H_f^{掳}\left(BrCl\left(g\right)\right)=14.78kJ/mol \end{gathered}$$

Therefore, the value is obtained as $4.78kJ/mol$.

(b)

Solve to obtain the value for $鈭�G_{rxn}^{掳}$ first 鈥�

$$\begin{gathered} 鈭�G_{rxn}^{掳}=\underset{}{\overset{}{鈭�}}{n}_p鈭�G_f^{掳}\left(product\right)-\underset{}{\overset{}{鈭�}}{n}_r鈭�G_f^{掳}\left(reac\tan t\right) \\ =\left[n鈭�G_f^{掳}\left(BrCl\left(g\right)\right)\right]-\left[n鈭�G_f^{掳}\left(C{l}_2\left(g\right)\right)+n鈭�G_f^{掳}\left(B{r}_2\left(g\right)\right)\right] \\ =\left(\left[2\left(-0.88\right)\right]-\left[1\left(0\right)+1\left(3.13\right)\right]\right)kJ/mol \\ 鈭�G_{rxn}^{掳}=-4.89kJ/mol \end{gathered}$$

Now, solve for $鈭�S_{rxn}^{掳}$ at$298K$ 鈥�

$$\begin{gathered} 鈭�G_{rxn}^{掳}=鈭�H_{rxn}^{掳}-T鈭�S_{rxn}^{掳} \\ T鈭�S_{rxn}^{掳}=鈭�H_{rxn}^{掳}-鈭�G_{rxn}^{掳} \\ 鈭�S_{rxn}^{掳}=\frac{鈭�H_{{}_{rxn}}^{掳}-鈭�G_{rxn}^{掳}}{T}=\frac{\left(2脳\left(-1.35kJ/mol\right)\right)-\left(-4.89kJ/mol\right)}{298K} \\ 鈭�S_{rxn}^{掳}=0.00735kJ/mol \\ =7.35J/mol路K \end{gathered}$$

Now manipulate the equation to solve for$S^{掳}$ of $BrCl\left(g\right)$ 鈥�

$$\begin{gathered} 鈭�S^{掳}=\underset{}{\overset{}{鈭�n_p}}{S}^{掳}\left(product\right)-\underset{}{\overset{}{鈭�}}{n}_r{S}^{掳}\left(reac\tan t\right) \\ 鈭�S^{掳}=\left[n{S}^{掳}\left(BrCl\left(g\right)\right)\right]-\left[n{S}^{掳}\left(B{r}_2\left(g\right)\right)+n{S}^{掳}\left(C{l}_2\left(g\right)\right)\right] \\ n{S}^{掳}\left(BrCl\left(g\right)\right)=鈭�S^{掳}+\left[n{S}^{掳}\left(B{r}_2\left(g\right)\right)+n{S}^{掳}\left(C{l}_2\left(g\right)\right)\right] \\ 2S^{掳}\left(BrCl\left(g\right)\right)=\left(7.35.J/mol路K\right)+\left(\left[1\left(223.0\right)+1\left(245.38\right)\right]\right)J/mol路K \\ {S}^{掳}\left(BrCl\left(g\right)\right)=\frac{475.73J/mol路K}{2} \\ {S}^{掳}\left(BrCl\left(g\right)\right)=237.9j/mol路K \end{gathered}$$

Therefore, the value is obtained as$237.9kJ/mol路K$ .