`DeltaH_(t)^(@)` for `CO_(2)(g)` and `H_(2)O(g)` are `-393.5, -110.5` and `-241.8 kJ"mol"^(-1)` respectively. The standard enthalpy change (in kJ) for the reaction.
`CO_(2)(g) + H_(2)(g) to CO(g) + H_(2)O(g)` is:

Options

A. `524.1`
B. `41.2`
C. `-262.5`
D. `-41.2`

Answer

To find the standard enthalpy change (ΔH°) for the reaction:

\[ \text{CO}_2(g) + \text{H}_2(g) \rightarrow \text{CO}(g) + \text{H}_2\text{O}(g) \]

we can use the standard enthalpy values provided for the substances involved in the reaction. The values given are:

- ΔH° for CO₂(g) = -393.5 kJ/mol
- ΔH° for H₂O(g) = -241.8 kJ/mol
- ΔH° for CO(g) = -110.5 kJ/mol

### Step 1: Write the reaction and identify the enthalpy changes

The reaction can be broken down as follows:

1. Reverse the formation of CO₂ to get it as a reactant:
\[ \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H° = +393.5 \text{ kJ/mol} \]

2. The formation of CO from its elements:
\[ \text{C}(s) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{CO}(g) \quad \Delta H° = -110.5 \text{ kJ/mol} \]

3. The formation of H₂O from its elements:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(g) \quad \Delta H° = -241.8 \text{ kJ/mol} \]

### Step 2: Combine the reactions

To find the ΔH° for the target reaction, we can combine the enthalpy changes of the reactions. We need to reverse the reaction for CO₂ and add the enthalpy changes:

\[ \Delta H° = \Delta H°(\text{reverse CO}_2) + \Delta H°(\text{formation of CO}) + \Delta H°(\text{formation of H}_2\text{O}) \]

### Step 3: Substitute the values

Substituting the values we have:

\[ \Delta H° = (+393.5 \text{ kJ/mol}) + (-110.5 \text{ kJ/mol}) + (-241.8 \text{ kJ/mol}) \]

### Step 4: Calculate the total ΔH°

Calculating the total:

\[ \Delta H° = 393.5 - 110.5 - 241.8 \]
\[ \Delta H° = 393.5 - 352.3 \]
\[ \Delta H° = 41.2 \text{ kJ/mol} \]

### Final Answer

The standard enthalpy change (ΔH°) for the reaction is:

\[ \Delta H° = +41.2 \text{ kJ/mol} \]

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