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Consider the chemical equations shown here:
[tex]\[
\begin{array}{l}
NO(g) + O_3(a) \rightarrow NO_2(g) + O_2(o) \quad \Delta H_1 = -1989 \, \text{kJ} \\
\frac{3}{2} O_2(g) \rightarrow O_3(Q) \quad \Delta H_2 = 142.3 \, \text{kJ} \\
O(Q) \rightarrow \frac{1}{2} O_2(Q) \quad \Delta H_3 = -247.5 \, \text{kJ}
\end{array}
\][/tex]

What is [tex]\(\Delta H\)[/tex] for the reaction shown below?

[tex]\[
\text{NO(g)} + \text{O}_3(\text{a}) \rightarrow \text{NO}_2(\text{g}) + \text{O}_2(\text{o})
\][/tex]



Answer :

GinnyAnswer
To determine the overall enthalpy change, [tex]\(\Delta H_{\text{total}}\)[/tex], for the reaction, you need to sum the individual enthalpy changes ([tex]\(\Delta H_1\)[/tex], [tex]\(\Delta H_2\)[/tex], and [tex]\(\Delta H_3\)[/tex]) provided for each step of the chemical process.

Here are the given reactions and their respective enthalpy changes:

1. [tex]\(NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g) \quad \Delta H_1 = -1989 \, \text{kJ}\)[/tex]
2. [tex]\(\frac{3}{2} O_2(g) \rightarrow O_3(g) \quad \Delta H_2 = 142.3 \, \text{kJ}\)[/tex]
3. [tex]\(O(g) \rightarrow \frac{1}{2} O_2(g) \quad \Delta H_3 = -247.5 \, \text{kJ}\)[/tex]

To find the overall enthalpy change for the combined reaction, we simply sum these enthalpy changes:

[tex]\[ \Delta H_{\text{total}} = \Delta H_1 + \Delta H_2 + \Delta H_3 \][/tex]

Plugging in the given values:

[tex]\[ \Delta H_{\text{total}} = -1989 \, \text{kJ} + 142.3 \, \text{kJ} + -247.5 \, \text{kJ} \][/tex]

When these values are summed, you get:

[tex]\[ \Delta H_{\text{total}} = -1989 + 142.3 - 247.5 = -2094.2 \, \text{kJ} \][/tex]

Thus, the overall enthalpy change [tex]\(\Delta H\)[/tex] for the reaction is:

[tex]\[ \Delta H_{\text{total}} = -2094.2 \, \text{kJ} \][/tex]

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