To determine the enthalpy change for the reaction [tex]\( H_2(g) + Br_2(g) \rightarrow 2 HBr(g) \)[/tex], we will use the given standard enthalpies of formation.
First, recall that the standard enthalpy change for a chemical reaction [tex]\(\Delta H_{reaction}\)[/tex] can be calculated using the standard enthalpies of formation [tex]\( \Delta H_f \)[/tex] of the reactants and products:
[tex]\[
\Delta H_{reaction} = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)}
\][/tex]
Given data from the table:
[tex]\[
\Delta H_f \text{(Br}_2(g)) = 30.907\ \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H_f \text{(HBr(g))} = -36.4\ \text{kJ/mol}
\][/tex]
[tex]\[
\Delta H_f \text{(H}_2(g)) = 0.0\ \text{kJ/mol}
\][/tex]
The balanced chemical equation is:
[tex]\[
H_2(g) + Br_2(g) \rightarrow 2 HBr(g)
\][/tex]
Now, let's calculate the total enthalpy of the products:
[tex]\[
\text{Total enthalpy of products} = 2 \times \Delta H_f \text{(HBr(g))}
\][/tex]
[tex]\[
= 2 \times -36.4\ \text{kJ/mol}
\][/tex]
[tex]\[
= -72.8\ \text{kJ/mol}
\][/tex]
Next, we calculate the total enthalpy of the reactants:
[tex]\[
\text{Total enthalpy of reactants} = \Delta H_f \text{(H}_2(g)) + \Delta H_f \text{(Br}_2(g))
\][/tex]
[tex]\[
= 0.0\ \text{kJ/mol} + 30.907\ \text{kJ/mol}
\][/tex]
[tex]\[
= 30.907\ \text{kJ/mol}
\][/tex]
Finally, we calculate the overall enthalpy change for the reaction:
[tex]\[
\Delta H_{reaction} = \text{Total enthalpy of products} - \text{Total enthalpy of reactants}
\][/tex]
[tex]\[
= -72.8\ \text{kJ/mol} - 30.907\ \text{kJ/mol}
\][/tex]
[tex]\[
= -103.707\ \text{kJ/mol}
\][/tex]
Therefore, the enthalpy change for the reaction [tex]\( H_2(g) + Br_2(g) \rightarrow 2 HBr(g) \)[/tex] is [tex]\(-103.707\ \text{kJ/mol}\)[/tex].
