Question

Bromine monochloride is formed from the elements:

$\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{=}\mathbf{-}{\mathbf{\text{1.35 kJ/mol∆G}}}_{\mathbf{r}\mathbf{x}\mathbf{n}}^{\mathbf{°}}\mathbf{\text{=-0.88 kJ/mol}}$

Calculate

(a) $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{°}}$and

(b) ${\mathit{S}}^{\mathbf{°}}$of$\mathbf{\text{BrCl(g)}}$

Short answer

1. The value of $∆H_f^{°}$ for$\text{BrCl(g)}$ is$14.78kJ/mol$ .
2. The value of $S^{°}$ for$\text{BrCl(g)}$ is $237.9kJ/mol·K$.

Step-by-step solution

Concept Introduction

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.

Calculation for ∆Hf°∆Hf°

(a)

The balanced equation is –

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

The values given are –

$$\begin{gathered} ∆H_{rxn}^{°}=-\text{1.35 kJ/mol} \\ ∆G_f^{°}=-\text{0.88 kJ/mol} \end{gathered}$$

Manipulate the reaction to solve for $∆H_f^{°}$of $\text{BrCl(g)}$ –

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

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

Calculation for S°

(b)

Solve to obtain the value for $∆G_{rxn}^{°}$ first –

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

Now, solve for $∆S_{rxn}^{°}$ at$298K$ –

$$\begin{gathered} ∆G_{rxn}^{°}=∆H_{rxn}^{°}-T∆S_{rxn}^{°} \\ T∆S_{rxn}^{°}=∆H_{rxn}^{°}-∆G_{rxn}^{°} \\ ∆S_{rxn}^{°}=\frac{∆H_{{}_{rxn}}^{°}-∆G_{rxn}^{°}}{T}=\frac{\left(2\times \left(-1.35kJ/mol\right)\right)-\left(-4.89kJ/mol\right)}{298K} \\ ∆S_{rxn}^{°}=0.00735kJ/mol \\ =7.35J/mol·K \end{gathered}$$

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

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

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