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Given the following enthalpies of reaction:
[tex]\[
\begin{array}{l}
C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g) \quad \Delta H = -1323 \, \text{kJ} \\
2CO(g) + O_2(g) \rightarrow 2CO_2(g) \quad \Delta H = -566 \, \text{kJ} \\
H_2O(g) \rightarrow H_2O(l) \quad \Delta H = +44 \, \text{kJ}
\end{array}
\][/tex]

Use Hess's Law to calculate [tex]\(\Delta H\)[/tex] for the reaction:
[tex]\[
C_2H_4(g) + 2O_2(g) \rightarrow 2CO(g) + 2H_2O(l)
\][/tex]

A. [tex]\(-801 \, \text{kJ}\)[/tex]
B. [tex]\(-757 \, \text{kJ}\)[/tex]
C. [tex]\(-845 \, \text{kJ}\)[/tex]
D. [tex]\(-1977 \, \text{kJ}\)[/tex]



Answer :

GinnyAnswer
To solve this problem using Hess's Law, we aim to combine the given reactions such that we achieve the target reaction:
[tex]\[ \text{C}_2\text{H}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO(g)} + 2\text{H}_2\text{O(g)} \][/tex]

We start by analyzing the given reactions:

1. [tex]\[ \text{C}_2\text{H}_4\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O (g)} \quad \Delta H = -1323 \text{ kJ} \][/tex]
2. [tex]\[ 2\text{CO(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} \quad \Delta H = -566 \text{ kJ} \][/tex]
3. [tex]\[ \text{H}_2\text{O (g)} \rightarrow \text{H}_2\text{O (g)} \quad \Delta H = +44 \text{ kJ} \][/tex] (Note: This reaction doesn't change anything, so it will not be used.)

Let's adjust the given reactions to form the target reaction:

- The first reaction provides the formation of [tex]\(\text{CO}_2\)[/tex] and [tex]\(\text{H}_2\text{O}\)[/tex] from [tex]\(\text{C}_2\text{H}_4\)[/tex] and [tex]\(\text{O}_2\)[/tex].
- The second reaction can be flipped to provide [tex]\(\text{CO}\)[/tex] from [tex]\(\text{CO}_2\)[/tex].

Reverse the second reaction:
[tex]\[ 2 \text{CO}_2\text{(g)} \rightarrow 2 \text{CO(g)} + \text{O}_2\text{(g)} \quad \Delta H = +566 \text{ kJ} \][/tex]

Now we have:
1. [tex]\[ \text{C}_2\text{H}_4\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O (g)} \quad \Delta H = -1323 \text{ kJ} \][/tex]
2. [tex]\[ 2\text{CO}_2\text{(g)} \rightarrow 2\text{CO(g)} + \text{O}_2\text{(g)} \quad \Delta H = +566 \text{ kJ} \][/tex]

Combine these two reactions:
[tex]\[ \text{C}_2\text{H}_4\text{(g)} + 3\text{O}_2\text{(g)} + 2\text{CO}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 2\text{H}_2\text{O (g)} + 2\text{CO(g)} + \text{O}_2\text{(g)} \][/tex]

Cancel out common terms on both sides:
[tex]\[ \text{C}_2\text{H}_4\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO(g)} + 2\text{H}_2\text{O (g)} + \text{O}_2\text{(g)} \][/tex]
[tex]\[ \text{C}_2\text{H}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO(g)} + 2\text{H}_2\text{O (g)} \][/tex]

Now, combine the enthalpies:
[tex]\[ \Delta H = -1323 \text{ kJ} + 566 \text{ kJ} = -757 \text{ kJ} \][/tex]

Thus, [tex]\(\Delta H\)[/tex] for the reaction [tex]\(\text{C}_2\text{H}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow 2\text{CO(g)} + 2\text{H}_2\text{O (g)}\)[/tex] is:
[tex]\[ \boxed{-757 \text{ kJ}} \][/tex]

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