Question

calculate the change in entropy of liquid bromine using the data below.
br₂(l) → br₂(g) δs° = +93 j/k
substance | s° (j/mol·k)
br₂(l) | ?
br₂(g) | 245
s°_br₂(l) = ? j/mol·k
enter either a + or - sign and the magnitude in your answer.

Explanation:

Step1: Recall the entropy change formula

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 know $\Delta S^\circ = +93 \, \text{J/K}$ and $S^\circ_{\text{Br}_2(\text{g})} = 245 \, \text{J/mol·K}$. Rearranging the formula gives $S^\circ_{\text{Br}_2(\text{l})} = S^\circ_{\text{Br}_2(\text{g})} - \Delta S^\circ$.

Step3: Substitute the values into the formula

Substitute $S^\circ_{\text{Br}_2(\text{g})} = 245$ and $\Delta S^\circ = 93$ into the formula: $S^\circ_{\text{Br}_2(\text{l})} = 245 - 93$.

Step4: Calculate the result

$245 - 93 = 152$. So the standard entropy of liquid bromine is $+152 \, \text{J/mol·K}$ (the positive sign indicates it's a standard entropy value, and since we're dealing with a substance in a state, the sign here is positive as entropy values are positive for substances).

Answer:

+152