High School

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Calculate [tex]\Delta H[/tex] for the following reaction:

[tex]S_{(s)} + 2 SO_3(\varphi) \rightarrow 3 SO_2(\underline{0}) \quad \Delta H=?[/tex]

Given the reference reactions below:

[tex]\begin{array}{ll}
S(\omega) + O_2(g) \rightarrow SO_2(g) & \Delta H = -296.0 \, \text{kJ/mol} \\
2 SO_2(v) + O_2(g) \rightarrow 2 SO_3(g) & \Delta H = -198.2 \, \text{kJ/mol}
\end{array}[/tex]

Choose the correct value for [tex]\Delta H[/tex]:

A. [tex]-97.8 \, \text{kJ/mol}[/tex]

B. [tex]-196.9 \, \text{kJ/mol}[/tex]

C. [tex]-494.2 \, \text{kJ/mol}[/tex]

D. [tex]-346.2 \, \text{kJ/mol}[/tex]

Answer :

We are given the following reference reactions:

[tex]$$
\text{Reaction 1:} \quad S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296.0\ \text{kJ/mol},
$$[/tex]

[tex]$$
\text{Reaction 2:} \quad 2\,SO_2(g) + O_2(g) \rightarrow 2\,SO_3(g) \quad \Delta H_2 = -198.2\ \text{kJ/mol}.
$$[/tex]

We need to calculate the enthalpy change for the reaction:

[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g) \quad \Delta H = ?
$$[/tex]

Step 1. Reverse Reaction 2

To obtain [tex]$2\,SO_3(g)$[/tex] on the left, we reverse Reaction 2. When a reaction is reversed, the sign of [tex]$\Delta H$[/tex] is also reversed. Thus, the reversed Reaction 2 becomes

[tex]$$
2\,SO_3(g) \rightarrow 2\,SO_2(g) + O_2(g) \quad \Delta H_{\text{rev2}} = +198.2\ \text{kJ/mol}.
$$[/tex]

Step 2. Add Reaction 1 and the Reversed Reaction 2

Now, write down the two reactions you will combine:

1. Reaction 1:
[tex]$$
S(s) + O_2(g) \rightarrow SO_2(g) \quad \Delta H_1 = -296.0\ \text{kJ/mol}
$$[/tex]
2. Reversed Reaction 2:
[tex]$$
2\,SO_3(g) \rightarrow 2\,SO_2(g) + O_2(g) \quad \Delta H_{\text{rev2}} = +198.2\ \text{kJ/mol}
$$[/tex]

Add these reactions together:

[tex]$$
\begin{array}{rcl}
S(s) + O_2(g) &\rightarrow& SO_2(g) \\
2\,SO_3(g) &\rightarrow& 2\,SO_2(g) + O_2(g)
\end{array}
$$[/tex]

When we add the left-hand sides and right-hand sides, we get:

[tex]$$
S(s) + O_2(g) + 2\,SO_3(g) \rightarrow SO_2(g) + 2\,SO_2(g) + O_2(g).
$$[/tex]

Step 3. Simplify the Overall Reaction

Notice that [tex]$O_2(g)$[/tex] appears on both sides and cancels out:

[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g).
$$[/tex]

This is exactly the reaction for which we want to find [tex]$\Delta H$[/tex].

Step 4. Calculate the Overall Enthalpy Change

The overall enthalpy change is the sum of the enthalpy changes of the individual steps:

[tex]$$
\Delta H = \Delta H_1 + \Delta H_{\text{rev2}}.
$$[/tex]

Substitute the values:

[tex]$$
\Delta H = (-296.0\ \text{kJ/mol}) + (198.2\ \text{kJ/mol}) = -97.8\ \text{kJ/mol}.
$$[/tex]

Thus, the enthalpy change for the reaction

[tex]$$
S(s) + 2\,SO_3(g) \rightarrow 3\,SO_2(g)
$$[/tex]

is

[tex]$$
\Delta H = -97.8\ \text{kJ/mol}.
$$[/tex]

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