High School

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

[tex]S_\omega + SO_2(0) + 2 O_2(0) \rightarrow 2 SO_2(0)[/tex]

[tex]\Delta H = ?[/tex]

Given the reference reactions below:

1. [tex]S_\omega + O_2(\hat{y}) \rightarrow SO_2(0)[/tex]
[tex]\Delta H = -296.0 \text{ kJ/mol}[/tex]

2. [tex]2 SO_2(\hat{y}) + O_2(y) \rightarrow 2 SO_3(\varphi)[/tex]
[tex]\Delta H = -198.2 \text{ kJ/mol}[/tex]

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

A. [tex]-494.2 \text{ kJ/mol}[/tex]
B. [tex]-97.8 \text{ kJ/mol}[/tex]
C. [tex]-346.2 \text{ kJ/mol}[/tex]

Answer :

We begin with the two reference reactions and their reaction enthalpies:

[tex]$$
\begin{array}{rcl}
\text{(1)}\quad S_\omega + O_2 &\to& SO_2\quad\quad \Delta H_1 = -296.0\ \text{kJ/mol}, \\[1mm]
\text{(2)}\quad 2\,SO_2 + O_2 &\to& 2\,SO_3\quad \Delta H_2 = -198.2\ \text{kJ/mol}.
\end{array}
$$[/tex]

Our target reaction is:

[tex]$$
S_\omega + SO_2 + 2\,O_2 \to 2\,SO_3.
$$[/tex]

Step 1. Notice that reaction (1) produces [tex]$SO_2$[/tex] starting from sulfur, and reaction (2) consumes [tex]$2\,SO_2$[/tex] to yield [tex]$2\,SO_3$[/tex]. In order to combine them appropriately, we add reaction (1) and reaction (2):

[tex]\[
\begin{array}{rcl}
S_\omega + O_2 &\to& SO_2 \quad\quad\quad\quad\quad\quad (\Delta H_1=-296.0\ \text{kJ/mol})\\[1mm]
2\,SO_2 + O_2 &\to& 2\,SO_3 \quad\quad\quad\quad (\Delta H_2=-198.2\ \text{kJ/mol})
\end{array}
\][/tex]

Step 2. Write the combined reaction by adding the left-hand sides and the right-hand sides:

[tex]\[
\begin{array}{rcl}
\text{Left side:} && S_\omega + O_2 + 2\,SO_2 + O_2 = S_\omega + 2\,SO_2 + 2\,O_2, \\[1mm]
\text{Right side:} && SO_2 + 2\,SO_3.
\end{array}
\][/tex]

Step 3. Notice that [tex]$SO_2$[/tex] appears on both sides. Subtract one mole of [tex]$SO_2$[/tex] from both sides to simplify the overall reaction:

[tex]$$
S_\omega + SO_2 + 2\,O_2 \to 2\,SO_3.
$$[/tex]

Step 4. Since enthalpy is a state function, the enthalpy change for the overall reaction is the sum of the enthalpy changes for the individual steps:

[tex]$$
\Delta H = \Delta H_1 + \Delta H_2 = (-296.0) + (-198.2) = -494.2\ \text{kJ/mol}.
$$[/tex]

Thus, the enthalpy change for the reaction is

[tex]$$
\boxed{-494.2 \text{ kJ/mol}}
$$[/tex]

This is the final answer.

Thank you for reading the article Calculate tex Delta H tex for the following reaction tex S omega SO 2 0 2 O 2 0 rightarrow 2 SO 2 0 tex. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!

Rewritten by : Jeany

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