Answered

Consider the following intermediate chemical equations:

[tex]\[
\begin{array}{ll}
C ( s )+ O _2( g ) \rightarrow CO _2( g ) & \Delta H_1 = -393.5 \, \text{kJ} \\
2 CO ( g )+ O _2( g ) \rightarrow 2 CO _2( g ) & \Delta H_2 = -566.0 \, \text{kJ} \\
2 H _2 O ( g ) \rightarrow 2 H _2( g )+ O _2( g ) & \Delta H_3 = 483.6 \, \text{kJ}
\end{array}
\][/tex]

The overall chemical equation is:

[tex]\[ C ( s ) + H _2 O ( g ) \rightarrow CO ( g ) + H _2( g ) \][/tex]

To calculate the final enthalpy of the overall chemical equation, which step must occur?

A. Reverse the first equation, and change the sign of the enthalpy. Then, add.

B. Reverse the second equation, and change the sign of the enthalpy. Then, add.

C. Multiply the first equation by three and triple the enthalpy. Then, add.

D. Divide the third equation by two, and double the enthalpy. Then, add.



Answer :

GinnyAnswer
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].