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Answer :
To find the standard enthalpy of formation for sulfur dioxide, [tex]\( SO_2(g) \)[/tex], we can use Hess's law, which states that the total enthalpy change for a chemical reaction is the same regardless of the pathway it takes.
We are given two reactions:
1. [tex]\( 2S(s) + 3O_2(g) \rightarrow 2SO_3(g) \)[/tex] with [tex]\(\Delta H^{\circ} = -790.4 \, \text{kJ}\)[/tex]
2. [tex]\( SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g) \)[/tex] with [tex]\(\Delta H^{\circ} = -99.1 \, \text{kJ}\)[/tex]
The goal is to find the standard enthalpy of formation for [tex]\( SO_2(g) \)[/tex], which corresponds to the reaction:
[tex]\[ S(s) + O_2(g) \rightarrow SO_2(g) \][/tex]
Here's how we can solve it:
1. The given reaction enthalpy for forming [tex]\( 2 \text{ moles of } SO_3\)[/tex] from sulfur is [tex]\(-790.4 \, \text{kJ}\)[/tex], which covers the entire process starting from elemental sulfur.
2. The reaction enthalpy for forming [tex]\( SO_3\)[/tex] from [tex]\( SO_2\)[/tex] is [tex]\(-99.1 \, \text{kJ}\)[/tex] per mole of [tex]\( SO_3\)[/tex]. For two moles, the enthalpy is:
[tex]\[ 2 \times (-99.1 \, \text{kJ}) = -198.2 \, \text{kJ} \][/tex]
3. According to Hess's law, the formation of [tex]\(2 \text{ moles of } SO_2\)[/tex] can be derived by subtracting the enthalpy change when [tex]\(2 \text{ moles of } SO_3\)[/tex] are formed from [tex]\( SO_2\)[/tex] from the enthalpy change when [tex]\( 2 \text{ moles of } SO_3\)[/tex] are formed from elemental sulfur.
[tex]\[ \Delta H_{2SO_2} = -790.4 \, \text{kJ} - (-198.2 \, \text{kJ}) \][/tex]
4. This gives:
[tex]\[ \Delta H_{2SO_2} = -790.4 \, \text{kJ} + 198.2 \, \text{kJ} = -592.2 \, \text{kJ} \][/tex]
5. Since this is for the formation of 2 moles of [tex]\( SO_2\)[/tex], the enthalpy for 1 mole of [tex]\( SO_2(g) \)[/tex] is:
[tex]\[ \Delta H_{SO_2} = \frac{-592.2 \, \text{kJ}}{2} = -296.1 \, \text{kJ} \][/tex]
So, the standard enthalpy of formation for sulfur dioxide, [tex]\( SO_2(g) \)[/tex], is [tex]\(-296.1 \, \text{kJ/mol}\)[/tex].
Thus, the correct answer is:
b) [tex]\(-296.1 \, \text{kJ/mol}\)[/tex]
We are given two reactions:
1. [tex]\( 2S(s) + 3O_2(g) \rightarrow 2SO_3(g) \)[/tex] with [tex]\(\Delta H^{\circ} = -790.4 \, \text{kJ}\)[/tex]
2. [tex]\( SO_2(g) + \frac{1}{2}O_2(g) \rightarrow SO_3(g) \)[/tex] with [tex]\(\Delta H^{\circ} = -99.1 \, \text{kJ}\)[/tex]
The goal is to find the standard enthalpy of formation for [tex]\( SO_2(g) \)[/tex], which corresponds to the reaction:
[tex]\[ S(s) + O_2(g) \rightarrow SO_2(g) \][/tex]
Here's how we can solve it:
1. The given reaction enthalpy for forming [tex]\( 2 \text{ moles of } SO_3\)[/tex] from sulfur is [tex]\(-790.4 \, \text{kJ}\)[/tex], which covers the entire process starting from elemental sulfur.
2. The reaction enthalpy for forming [tex]\( SO_3\)[/tex] from [tex]\( SO_2\)[/tex] is [tex]\(-99.1 \, \text{kJ}\)[/tex] per mole of [tex]\( SO_3\)[/tex]. For two moles, the enthalpy is:
[tex]\[ 2 \times (-99.1 \, \text{kJ}) = -198.2 \, \text{kJ} \][/tex]
3. According to Hess's law, the formation of [tex]\(2 \text{ moles of } SO_2\)[/tex] can be derived by subtracting the enthalpy change when [tex]\(2 \text{ moles of } SO_3\)[/tex] are formed from [tex]\( SO_2\)[/tex] from the enthalpy change when [tex]\( 2 \text{ moles of } SO_3\)[/tex] are formed from elemental sulfur.
[tex]\[ \Delta H_{2SO_2} = -790.4 \, \text{kJ} - (-198.2 \, \text{kJ}) \][/tex]
4. This gives:
[tex]\[ \Delta H_{2SO_2} = -790.4 \, \text{kJ} + 198.2 \, \text{kJ} = -592.2 \, \text{kJ} \][/tex]
5. Since this is for the formation of 2 moles of [tex]\( SO_2\)[/tex], the enthalpy for 1 mole of [tex]\( SO_2(g) \)[/tex] is:
[tex]\[ \Delta H_{SO_2} = \frac{-592.2 \, \text{kJ}}{2} = -296.1 \, \text{kJ} \][/tex]
So, the standard enthalpy of formation for sulfur dioxide, [tex]\( SO_2(g) \)[/tex], is [tex]\(-296.1 \, \text{kJ/mol}\)[/tex].
Thus, the correct answer is:
b) [tex]\(-296.1 \, \text{kJ/mol}\)[/tex]
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