Let's solve the problem step-by-step to find the enthalpy for the overall chemical equation [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex].
### Given Equations and their Enthalpies
1. [tex]\( C(s) + O_2(g) \rightarrow CO_2(g) \hspace{20pt} \Delta H_1 = -393.5 \, \text{kJ} \)[/tex]
2. [tex]\( 2 \, CO(g) + O_2(g) \rightarrow 2 \, CO_2(g) \hspace{20pt} \Delta H_2 = -566.0 \, \text{kJ} \)[/tex]
3. [tex]\( 2 \, H_2O(g) \rightarrow 2 \, H_2(g) + O_2(g) \hspace{20pt} \Delta H_3 = 483.6 \, \text{kJ} \)[/tex]
### Step 1: Reverse the First Equation
To match the final equation, we need carbon monoxide as a product, hence we reverse the first reaction.
[tex]\[ CO_2(g) \rightarrow C(s) + O_2(g) \][/tex]
The enthalpy change of this reversed reaction is
[tex]\[ \Delta H'_1 = +393.5 \, \text{kJ} \][/tex]
### Step 2: Reverse the Second Equation
We need carbon monoxide as a product, so we reverse the second reaction.
[tex]\[ 2 \, CO_2(g) \rightarrow 2 \, CO(g) + O_2(g) \][/tex]
The enthalpy change of this reversed reaction is
[tex]\[ \Delta H'_2 = +566.0 \, \text{kJ} \][/tex]
### Step 3: Adjust the Third Equation
We need [tex]\( H_2 \)[/tex] and [tex]\( H_2O \)[/tex] in small amounts; hence, we divide the third reaction by 2.
[tex]\[ H_2O(g) \rightarrow H_2(g) + \frac{1}{2} O_2(g) \][/tex]
The enthalpy for this adjusted equation is half of the original:
[tex]\[ \Delta H'_3 = \frac{483.6 \, \text{kJ}}{2} = 241.8 \, \text{kJ} \][/tex]
### Step 4: Summing Up the Enthalpies
Now, we sum the enthalpies of the adjusted equations to find the enthalpy of the overall reaction:
[tex]\[ \Delta H_{\text{total}} = \Delta H'_1 + \Delta H'_2 + \Delta H'_3 \][/tex]
[tex]\[ \Delta H_{\text{total}} = 393.5 \, \text{kJ} + 566.0 \, \text{kJ} + 241.8 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{total}} = 1201.3 \, \text{kJ} \][/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]\( \Delta H_{\text{total}} = 1201.3 \, \text{kJ} \)[/tex].
