Answer:
The answer to your question is given below.
Explanation:
From the question given above, the following data were obtained:
Br₂ (l) —> Br₂(g)
Enthalpy change (ΔH) = 30.91 KJ/mol
Entropy change (ΔS) = 93.3 J/mol·K
Boiling temperature (T) =?
Next, we shall convert 30.91 KJ/mol to J/mol. This can be obtained as follow:
1 KJ/mol = 1000 J/mol
Therefore,
30.91 KJ/mol = 30.91 × 1000
30.91 KJ/mol = 30910 J/mol
Thus, 30.91 KJ/mol is equivalent to 30910 J/mol.
Finally, we shall determine the boiling temperature of bromine. This can be obtained as follow:
Enthalpy change (ΔH) = 30910 J/mol
Entropy change (ΔS) = 93.3 J/mol·K
Boiling temperature (T) =?
ΔS = ΔH / T
93.3 = 30910 / T
Cross multiply
93.3 × T = 30910
Divide both side by 93.3
T = 30910 / 93.3
T = 331.29 K
Thus, the boiling temperature of bromine is 331.29 K
