Question

calculate the change in entropy of liquid bromine using the data below.\\(\ce{br2(l) -> br2(g)} \quad \delta s^\circ = +93\\,{j/k}\\)\
\

$$\begin{tabular}{|c|c|} \\hline substance & \\(s^\\circ\\,{(j/mol\\cdot k)}\\) \\\\ \\hline \\(\\ce{br2(l)}\\) & \\(\\{?\\}\\) \\\\ \\hline \\(\\ce{br2(g)}\\) & 245 \\\\ \\hline \\end{tabular}$$

\
\\(s^\circ_{\ce{br2(l)}} = \\{?\\}\\,{j/mol\cdot k}\\)\
enter either a + or - sign and the magnitude in your answer.

Explanation:

Step1: Recall the formula for entropy change

The standard entropy change of a reaction, $\Delta S^{\circ}$, is given by the sum of the standard entropies of the products minus the sum of the standard entropies of the reactants. For the reaction $\text{Br}_2(\text{l})
ightarrow \text{Br}_2(\text{g})$, the formula is $\Delta S^{\circ} = S^{\circ}_{\text{Br}_2(\text{g})} - S^{\circ}_{\text{Br}_2(\text{l})}$.

Step2: Rearrange the formula to solve for $S^{\circ}_{\text{Br}_2(\text{l})}$

We can rearrange the formula to $S^{\circ}_{\text{Br}_2(\text{l})} = S^{\circ}_{\text{Br}_2(\text{g})} - \Delta S^{\circ}$.

Step3: Substitute the known values

We know that $S^{\circ}_{\text{Br}_2(\text{g})} = 245 \, \text{J/mol·K}$ and $\Delta S^{\circ} = +93 \, \text{J/K}$. Substituting these values into the formula, we get $S^{\circ}_{\text{Br}_2(\text{l})} = 245 - 93$.

Step4: Calculate the result

Performing the subtraction, $245 - 93 = 152$. So, $S^{\circ}_{\text{Br}_2(\text{l})} = +152 \, \text{J/mol·K}$ (the positive sign indicates the magnitude as entropy values are positive for these substances in standard state).

Answer:

+152