To calculate the standard enthalpy change for the reaction, we need to use the standard enthalpies of formation ([tex]\( \Delta H_f^\circ \)[/tex]) provided for each substance involved in the balanced chemical equation:
[tex]\[
2A + B \longrightarrow 2C + 2D
\][/tex]
The standard enthalpy change of the reaction ([tex]\( \Delta H_{rxn}^\circ \)[/tex]) can be calculated using the enthalpies of formation of the reactants and products according to the following formula:
[tex]\[
\Delta H_{rxn}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})
\][/tex]
This involves summing the enthalpies of formation of the products and reactants, weighted by their stoichiometric coefficients.
Given:
- [tex]\( \Delta H_f^\circ(A) = -259 \)[/tex] kJ/mol
- [tex]\( \Delta H_f^\circ(B) = -397 \)[/tex] kJ/mol
- [tex]\( \Delta H_f^\circ(C) = 201 \)[/tex] kJ/mol
- [tex]\( \Delta H_f^\circ(D) = -481 \)[/tex] kJ/mol
The balanced equation tells us the stoichiometric coefficients:
- For product [tex]\( C \)[/tex]: coefficient is 2
- For product [tex]\( D \)[/tex]: coefficient is 2
- For reactant [tex]\( A \)[/tex]: coefficient is 2
- For reactant [tex]\( B \)[/tex]: coefficient is 1
Now plug these values into the formula:
[tex]\[
\Delta H_{rxn}^\circ = [2 \Delta H_f^\circ(C) + 2 \Delta H_f^\circ(D)] - [2 \Delta H_f^\circ(A) + \Delta H_f^\circ(B)]
\][/tex]
Substitute the given values:
[tex]\[
\Delta H_{rxn}^\circ = [2 \times 201 + 2 \times (-481)] - [2 \times (-259) + (-397)]
\][/tex]
Simplify the expression step-by-step:
[tex]\[
\Delta H_{rxn}^\circ = [402 + (-962)] - [(-518) + (-397)]
\][/tex]
[tex]\[
\Delta H_{rxn}^\circ = [402 - 962] - [-518 - 397]
\][/tex]
[tex]\[
\Delta H_{rxn}^\circ = -560 - (-915)
\][/tex]
[tex]\[
\Delta H_{rxn}^\circ = -560 + 915
\][/tex]
[tex]\[
\Delta H_{rxn}^\circ = 355
\][/tex]
Therefore, the standard enthalpy change for the reaction is:
[tex]\[
\Delta H_{rxn}^\circ = 355 \text{ kJ}
\][/tex]
