To calculate the final enthalpy change ([tex]\(\Delta H_{\text{total}}\)[/tex]) for the overall chemical equation [tex]\(C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)\)[/tex], we need to manipulate the given intermediate equations correctly. Here are the detailed steps:
1. Reverse the first equation:
The given equation is [tex]\(C(s) + O_2(g) \rightarrow CO_2(g)\)[/tex] with [tex]\(\Delta H_1 = -393.5 \, \text{kJ}\)[/tex].
Reversing this equation gives:
[tex]\(CO_2(g) \rightarrow C(s) + O_2(g)\)[/tex], and the enthalpy change also reverses its sign:
[tex]\(\Delta H_1' = 393.5 \, \text{kJ}\)[/tex].
2. Reverse the second equation:
The given equation is [tex]\(2 CO(g) + O_2(g) \rightarrow 2 CO_2(g)\)[/tex] with [tex]\(\Delta H_2 = -566.0 \, \text{kJ}\)[/tex].
Reversing this equation gives:
[tex]\(2 CO_2(g) \rightarrow 2 CO(g) + O_2(g)\)[/tex], and the enthalpy change also reverses its sign:
[tex]\(\Delta H_2' = 566.0 \, \text{kJ}\)[/tex].
3. Divide the third equation by two:
The given equation is [tex]\(2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g)\)[/tex] with [tex]\(\Delta H_3 = 483.6 \, \text{kJ}\)[/tex].
Dividing this equation by two gives:
[tex]\(H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g)\)[/tex], and the enthalpy change is also halved:
[tex]\(\Delta H_3' = \frac{1}{2} \times 483.6 \, \text{kJ} = 241.8 \, \text{kJ}\)[/tex].
4. Final enthalpy change for the overall reaction:
Now we need to sum up the enthalpy changes of the reversed and modified equations to find the total enthalpy change:
[tex]\[
\Delta H_{\text{total}} = \Delta H_1' + \Delta H_2' + \Delta H_3'
\][/tex]
Substituting the values we get:
[tex]\[
\Delta H_{\text{total}} = 393.5 \, \text{kJ} + 566.0 \, \text{kJ} + 241.8 \, \text{kJ} = 1201.3 \, \text{kJ}
\][/tex]
Therefore, the final enthalpy change for the overall chemical equation [tex]\(C(s) + H_2O(g) \rightarrow CO(g) + H_2(g)\)[/tex] is [tex]\(1201.3 \, \text{kJ}\)[/tex].
