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Answer :
To find the standard enthalpy change for the reaction [tex]\( SO_2(g) + \frac{1}{2} O_2(g) \rightarrow SO_3(g) \)[/tex], we will use the enthalpies of formation provided in the table.
1. Identify the standard enthalpy of formation values:
- For [tex]\( SO_2(g) \)[/tex], [tex]\(\Delta H_f^\circ = -296.1 \, \text{kJ/mol}\)[/tex]
- For [tex]\( SO_3(g) \)[/tex], [tex]\(\Delta H_f^\circ = -395.2 \, \text{kJ/mol}\)[/tex]
- For [tex]\( O_2(g) \)[/tex], [tex]\(\Delta H_f^\circ = 0.0 \, \text{kJ/mol}\)[/tex]
2. Write the expression for the enthalpy change of the reaction ([tex]\(\Delta H^\circ\)[/tex]):
[tex]\[
\Delta H^\circ = \text{(sum of the standard enthalpies of formation of products)} - \text{(sum of the standard enthalpies of formation of reactants)}
\][/tex]
3. Substitute the values into this expression:
[tex]\[
\Delta H^\circ = \left[ \Delta H_f^\circ (SO_3) \right] - \left[ \Delta H_f^\circ (SO_2) + \frac{1}{2} \Delta H_f^\circ (O_2) \right]
\][/tex]
4. Plug in the given values:
[tex]\[
\Delta H^\circ = \left[ -395.2 \right] - \left[ -296.1 + \frac{1}{2} \times 0.0 \right]
\][/tex]
5. Simplify the expression:
[tex]\[
\Delta H^\circ = -395.2 - (-296.1)
\][/tex]
6. Calculate the value:
[tex]\[
\Delta H^\circ = -395.2 + 296.1 = -99.1 \, \text{kJ/mol}
\][/tex]
Hence, the standard enthalpy change for the reaction is [tex]\(-99.1 \, \text{kJ/mol}_{rxn}\)[/tex].
The correct answer is:
(B) [tex]\(-99.1 \, \text{kJ/mol}_{rxn}\)[/tex]
1. Identify the standard enthalpy of formation values:
- For [tex]\( SO_2(g) \)[/tex], [tex]\(\Delta H_f^\circ = -296.1 \, \text{kJ/mol}\)[/tex]
- For [tex]\( SO_3(g) \)[/tex], [tex]\(\Delta H_f^\circ = -395.2 \, \text{kJ/mol}\)[/tex]
- For [tex]\( O_2(g) \)[/tex], [tex]\(\Delta H_f^\circ = 0.0 \, \text{kJ/mol}\)[/tex]
2. Write the expression for the enthalpy change of the reaction ([tex]\(\Delta H^\circ\)[/tex]):
[tex]\[
\Delta H^\circ = \text{(sum of the standard enthalpies of formation of products)} - \text{(sum of the standard enthalpies of formation of reactants)}
\][/tex]
3. Substitute the values into this expression:
[tex]\[
\Delta H^\circ = \left[ \Delta H_f^\circ (SO_3) \right] - \left[ \Delta H_f^\circ (SO_2) + \frac{1}{2} \Delta H_f^\circ (O_2) \right]
\][/tex]
4. Plug in the given values:
[tex]\[
\Delta H^\circ = \left[ -395.2 \right] - \left[ -296.1 + \frac{1}{2} \times 0.0 \right]
\][/tex]
5. Simplify the expression:
[tex]\[
\Delta H^\circ = -395.2 - (-296.1)
\][/tex]
6. Calculate the value:
[tex]\[
\Delta H^\circ = -395.2 + 296.1 = -99.1 \, \text{kJ/mol}
\][/tex]
Hence, the standard enthalpy change for the reaction is [tex]\(-99.1 \, \text{kJ/mol}_{rxn}\)[/tex].
The correct answer is:
(B) [tex]\(-99.1 \, \text{kJ/mol}_{rxn}\)[/tex]
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