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Sulfur reacts with oxygen to form sulfur dioxide [tex]$SO_2(g)$[/tex], with [tex]\Delta H_f = -296.8 \, \text{kJ/mol}[/tex] according to the equation:

[tex] S(s) + O_2(g) \rightarrow SO_2(g) [/tex]

What is the enthalpy change for the reaction?

Use the equation:

[tex] \Delta H_{r \times n} = \sum (\Delta H_{\text{f, products}}) - \sum (\Delta H_{\text{f, reactants}}) [/tex]

A. [tex] -593.6 \, \text{kJ} [/tex]
B. [tex] -296.8 \, \text{kJ} [/tex]
C. [tex] 296.8 \, \text{kJ} [/tex]
D. [tex] 593.6 \, \text{kJ} [/tex]


Sagot :

To determine the enthalpy change for the reaction, we will use the given equation and the provided values.

[tex]\[ S(s) + O_2(g) \rightarrow SO_2(g) \][/tex]

We know the formula to calculate the enthalpy change for the reaction:

[tex]\[ \Delta H_{rxn} = \sum\left(\Delta H_{f, \text{products}}\right) - \sum\left(\Delta H_{f, \text{reactants}}\right) \][/tex]

Step-by-Step Breakdown:

1. Identify the ΔH_f values:
- For the product [tex]\(SO_2(g)\)[/tex], the enthalpy of formation [tex]\(\Delta H_f = -296.8 \text{ kJ/mol}\)[/tex] as given.
- For the reactants, sulfur (S(s)) and oxygen (O_2(g)) are in their standard states. The enthalpies of formation for elements in their standard states are zero:
[tex]\[ \Delta H_f(S(s)) = 0 \text{ kJ/mol} \][/tex]
[tex]\[ \Delta H_f(O_2(g)) = 0 \text{ kJ/mol} \][/tex]

2. Calculate the sum of the enthalpies of formation for the products:
[tex]\[ \sum\left(\Delta H_{f, \text{products}}\right) = \Delta H_f(SO_2(g)) = -296.8 \text{ kJ/mol} \][/tex]

3. Calculate the sum of the enthalpies of formation for the reactants:
[tex]\[ \sum\left(\Delta H_{f, \text{reactants}}\right) = \Delta H_f(S(s)) + \Delta H_f(O_2(g)) = 0 + 0 = 0 \text{ kJ/mol} \][/tex]

4. Plug these values into the enthalpy change formula:
[tex]\[ \Delta H_{rxn} = \sum\left(\Delta H_{f, \text{products}}\right) - \sum\left(\Delta H_{f, \text{reactants}}\right) \][/tex]
[tex]\[ \Delta H_{rxn} = -296.8 \text{ kJ/mol} - 0 \text{ kJ/mol} \][/tex]
[tex]\[ \Delta H_{rxn} = -296.8 \text{ kJ/mol} \][/tex]

Hence, the enthalpy change for the reaction is [tex]\(-296.8 \text{ kJ/mol}\)[/tex].

So, the correct answer is:

[tex]\[ \boxed{-296.8 \text{ kJ}} \][/tex]