What is ∆H0 for this reaction?
Br2(g) → Br2(l)
(∆Hf0 Br2(g) = 30.91 kJ/mol)
1.546 × 101 kJ
–1.546 × 101 kJ
3.091 × 101 kJ
–3.091 × 101 kJ
Br2(g) → Br2(l)
dHorxn = (dHof of products) - (dHof of reactants)
dHo formation Br2(g) = 30.91 kJ/mol
dHo formation Br2(l) = 0 kJ/mol
dHorxn = ?
I obtained the 30.91 from tables I have in a text. You could have looked this up in your text or notes.
To determine ∆H0 for the reaction, we need to consider the difference in enthalpy between the products and the reactant.
Given:
∆Hf0 Br2(g) = 30.91 kJ/mol
The reaction is:
Br2(g) → Br2(l)
Since the reaction involves a phase change from gas (g) to liquid (l), we can use the enthalpy of vaporization (∆Hvap) to determine the ∆H0.
The ∆H0 for the reaction is the sum of the enthalpy of vaporization (∆Hvap) and the enthalpy of formation (∆Hf0) of the liquid phase of bromine (Br2(l)):
∆H0 = ∆Hvap + ∆Hf0
However, the enthalpy of vaporization (∆Hvap) is not given. Therefore, we cannot calculate the ∆H0 for the reaction.
To determine the standard enthalpy change (∆H0) for the given reaction, we need to consider the difference in standard enthalpies of formation (∆Hf0) for the products and reactants.
For the reaction Br2(g) → Br2(l), the change in state from gaseous Br2 to liquid Br2 suggests that heat is being released (exothermic reaction). Therefore, the standard enthalpy change (∆H0) for this reaction will be negative.
Given that ∆Hf0 Br2(g) = 30.91 kJ/mol, we can compare this value with the remaining options:
1.546 × 101 kJ: This does not match the expected negative sign.
–1.546 × 101 kJ: This matches the signs, but the magnitude does not match the given ∆Hf0 value.
3.091 × 101 kJ: This does not match the expected negative sign.
–3.091 × 101 kJ: This matches the signs and magnitude of ∆Hf0 Br2(g).
Therefore, the correct answer is –3.091 × 101 kJ, which represents the standard enthalpy change for the reaction Br2(g) → Br2(l).