Answer :
To find the standard enthalpy change ([tex]\(\Delta H^{\circ}\)[/tex]) for the reaction [tex]\(2 \text{S}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{SO}_3(g)\)[/tex], you can use Hess's Law. This law states that the total enthalpy change of a reaction is the sum of the enthalpy changes of its individual steps.
Given Reactions:
1. [tex]\( \text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \)[/tex] with [tex]\(\Delta H^{\circ} = -296.1 \, \text{kJ}\)[/tex]
2. [tex]\( 2 \text{SO}_3(g) \rightarrow 2 \text{SO}_2(g) + \text{O}_2(g) \)[/tex] with [tex]\(\Delta H^{\circ} = 198.2 \, \text{kJ}\)[/tex]
Steps to Solve:
1. Modify the First Reaction:
- The first reaction needs to be multiplied by 2 to match the number of sulfur atoms in our target reaction:
[tex]\[
2(\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g))
\][/tex]
- This gives:
[tex]\[
2\text{S}(s) + 2\text{O}_2(g) \rightarrow 2\text{SO}_2(g)
\][/tex]
- Modified [tex]\(\Delta H^{\circ} = 2 \times (-296.1 \, \text{kJ}) = -592.2 \, \text{kJ}\)[/tex]
2. Reverse the Second Reaction:
- Reverse the second reaction to form [tex]\( \text{SO}_3 \)[/tex] from [tex]\( \text{SO}_2 \)[/tex]:
[tex]\[
2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)
\][/tex]
- Reverse the sign of [tex]\(\Delta H^{\circ}\)[/tex]:
[tex]\[
\Delta H^{\circ} = -198.2 \, \text{kJ}
\][/tex]
3. Add the Reactions:
- Combine the modified first reaction and the reversed second reaction:
[tex]\[
(2\text{S}(s) + 2\text{O}_2(g) \rightarrow 2\text{SO}_2(g)) + (2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g))
\][/tex]
- The [tex]\(\text{SO}_2\)[/tex] cancels out, leaving:
[tex]\[
2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)
\][/tex]
4. Calculate Total [tex]\(\Delta H\)[/tex]:
[tex]\[
\Delta H^{\circ}_{\text{total}} = -592.2 \, \text{kJ} + (-198.2 \, \text{kJ}) = -790.4 \, \text{kJ}
\][/tex]
Therefore, the standard enthalpy change for the reaction is [tex]\(-790.4 \, \text{kJ}\)[/tex].
The correct answer is B) -790.4 kJ.
Given Reactions:
1. [tex]\( \text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \)[/tex] with [tex]\(\Delta H^{\circ} = -296.1 \, \text{kJ}\)[/tex]
2. [tex]\( 2 \text{SO}_3(g) \rightarrow 2 \text{SO}_2(g) + \text{O}_2(g) \)[/tex] with [tex]\(\Delta H^{\circ} = 198.2 \, \text{kJ}\)[/tex]
Steps to Solve:
1. Modify the First Reaction:
- The first reaction needs to be multiplied by 2 to match the number of sulfur atoms in our target reaction:
[tex]\[
2(\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g))
\][/tex]
- This gives:
[tex]\[
2\text{S}(s) + 2\text{O}_2(g) \rightarrow 2\text{SO}_2(g)
\][/tex]
- Modified [tex]\(\Delta H^{\circ} = 2 \times (-296.1 \, \text{kJ}) = -592.2 \, \text{kJ}\)[/tex]
2. Reverse the Second Reaction:
- Reverse the second reaction to form [tex]\( \text{SO}_3 \)[/tex] from [tex]\( \text{SO}_2 \)[/tex]:
[tex]\[
2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g)
\][/tex]
- Reverse the sign of [tex]\(\Delta H^{\circ}\)[/tex]:
[tex]\[
\Delta H^{\circ} = -198.2 \, \text{kJ}
\][/tex]
3. Add the Reactions:
- Combine the modified first reaction and the reversed second reaction:
[tex]\[
(2\text{S}(s) + 2\text{O}_2(g) \rightarrow 2\text{SO}_2(g)) + (2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g))
\][/tex]
- The [tex]\(\text{SO}_2\)[/tex] cancels out, leaving:
[tex]\[
2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)
\][/tex]
4. Calculate Total [tex]\(\Delta H\)[/tex]:
[tex]\[
\Delta H^{\circ}_{\text{total}} = -592.2 \, \text{kJ} + (-198.2 \, \text{kJ}) = -790.4 \, \text{kJ}
\][/tex]
Therefore, the standard enthalpy change for the reaction is [tex]\(-790.4 \, \text{kJ}\)[/tex].
The correct answer is B) -790.4 kJ.
