Answer :
To calculate the final enthalpy of the overall chemical equation [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex], we need to consider the following steps:
### Step-by-Step Solution
1. Given Intermediate Reactions and Their Enthalpies:
[tex]\[ \begin{array}{ll} \text{(1)} \quad C(s) + O_2(g) \rightarrow CO_2(g) & \Delta H_1 = -393.5 \text{ kJ} \\ \text{(2)} \quad 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) & \Delta H_2 = -566.0 \text{ kJ} \\ \text{(3)} \quad 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) & \Delta H_3 = 483.6 \text{ kJ} \end{array} \][/tex]
2. Reverse the First Reaction:
Since we need [tex]\( C(s) \)[/tex] to appear as a product in the final equation, we reverse the first reaction:
[tex]\[ \text{Reverse of (1)} \quad CO_2(g) \rightarrow C(s) + O_2(g) \][/tex]
When we reverse a reaction, the sign of the enthalpy change also reverses:
[tex]\[ \Delta H_1 = +393.5 \text{ kJ} \][/tex]
3. Hess's Law:
According to Hess's Law, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual steps.
4. Match the Final Reaction:
We aim to derive [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex] from these intermediate reactions.
- Step 1: Use the reversed first reaction:
[tex]\[ CO_2(g) \rightarrow C(s) + O_2(g) \quad (\Delta H = +393.5 \text{ kJ}) \][/tex]
- Step 2: Use the second reaction as is:
[tex]\[ 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \quad (\Delta H = -566.0 \text{ kJ}) \][/tex]
- Step 3: Use the third reaction as is:
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) \quad (\Delta H = 483.6 \text{ kJ}) \][/tex]
5. Combine Equations:
- Reversed first reaction gives:
[tex]\[ CO_2(g) \rightarrow C(s) + O_2(g) \quad (\Delta H = 393.5 \text{ kJ}) \][/tex]
- Second reaction:
[tex]\[ CO(g) \rightarrow 0.5 O_2(g) + CO_2(g) \quad (\frac{-566.0}{2} = -283.0 \text{ kJ}) \][/tex]
- Third reaction:
[tex]\[ H_2O(g) \rightarrow H_2(g) + 0.5 O_2(g) \quad (\frac{483.6}{2} = 241.8 \text{ kJ}) \][/tex]
6. Sum of Enthalpies:
[tex]\[ \Delta H_{\text{first reversed}} + \Delta H_{\text{second}} + \Delta H_{\text{third}} = 393.5 \text{ kJ} + (-283.0 \text{ kJ}) + 241.8 \text{ kJ} \][/tex]
7. Calculate the Final Enthalpy:
[tex]\[ \Delta H_{\text{overall}} = 393.5 \text{ kJ} + (-283.0 \text{ kJ}) + 241.8 \text{ kJ} = 311.1 \text{ kJ} \][/tex]
Therefore, the enthalpy change for the overall reaction [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex] is [tex]\( 311.1 \text{ kJ} \)[/tex].
### Step-by-Step Solution
1. Given Intermediate Reactions and Their Enthalpies:
[tex]\[ \begin{array}{ll} \text{(1)} \quad C(s) + O_2(g) \rightarrow CO_2(g) & \Delta H_1 = -393.5 \text{ kJ} \\ \text{(2)} \quad 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) & \Delta H_2 = -566.0 \text{ kJ} \\ \text{(3)} \quad 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) & \Delta H_3 = 483.6 \text{ kJ} \end{array} \][/tex]
2. Reverse the First Reaction:
Since we need [tex]\( C(s) \)[/tex] to appear as a product in the final equation, we reverse the first reaction:
[tex]\[ \text{Reverse of (1)} \quad CO_2(g) \rightarrow C(s) + O_2(g) \][/tex]
When we reverse a reaction, the sign of the enthalpy change also reverses:
[tex]\[ \Delta H_1 = +393.5 \text{ kJ} \][/tex]
3. Hess's Law:
According to Hess's Law, the enthalpy change for the overall reaction is the sum of the enthalpy changes of the individual steps.
4. Match the Final Reaction:
We aim to derive [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex] from these intermediate reactions.
- Step 1: Use the reversed first reaction:
[tex]\[ CO_2(g) \rightarrow C(s) + O_2(g) \quad (\Delta H = +393.5 \text{ kJ}) \][/tex]
- Step 2: Use the second reaction as is:
[tex]\[ 2 CO(g) + O_2(g) \rightarrow 2 CO_2(g) \quad (\Delta H = -566.0 \text{ kJ}) \][/tex]
- Step 3: Use the third reaction as is:
[tex]\[ 2 H_2O(g) \rightarrow 2 H_2(g) + O_2(g) \quad (\Delta H = 483.6 \text{ kJ}) \][/tex]
5. Combine Equations:
- Reversed first reaction gives:
[tex]\[ CO_2(g) \rightarrow C(s) + O_2(g) \quad (\Delta H = 393.5 \text{ kJ}) \][/tex]
- Second reaction:
[tex]\[ CO(g) \rightarrow 0.5 O_2(g) + CO_2(g) \quad (\frac{-566.0}{2} = -283.0 \text{ kJ}) \][/tex]
- Third reaction:
[tex]\[ H_2O(g) \rightarrow H_2(g) + 0.5 O_2(g) \quad (\frac{483.6}{2} = 241.8 \text{ kJ}) \][/tex]
6. Sum of Enthalpies:
[tex]\[ \Delta H_{\text{first reversed}} + \Delta H_{\text{second}} + \Delta H_{\text{third}} = 393.5 \text{ kJ} + (-283.0 \text{ kJ}) + 241.8 \text{ kJ} \][/tex]
7. Calculate the Final Enthalpy:
[tex]\[ \Delta H_{\text{overall}} = 393.5 \text{ kJ} + (-283.0 \text{ kJ}) + 241.8 \text{ kJ} = 311.1 \text{ kJ} \][/tex]
Therefore, the enthalpy change for the overall reaction [tex]\( C(s) + H_2O(g) \rightarrow CO(g) + H_2(g) \)[/tex] is [tex]\( 311.1 \text{ kJ} \)[/tex].
