The `DeltaH_(f)^(@)` for `CO_(2)(g)`, CO(g) and `H_(2)O(g)` are `-395.5,-110.5` and `-241.8" kJ" mol^(-1)` respectively. The standard enthalpy change in (in kJ) for the reaction
`CO_(2)(g)+H_(2)(g)toCO(g)+H_(2)O(g)` is

Options

A. 524 .1
B. 41 .2
C. `-262 .5`
D. `-41.2`

Answer

To calculate the standard enthalpy change for the reaction:

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

we will use the standard enthalpy of formation values provided for each substance involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Enthalpy of Formation Values**:
- For \( \text{CO}_2(g) \): \( \Delta H_f^\circ = -395.5 \, \text{kJ/mol} \)
- For \( \text{CO}(g) \): \( \Delta H_f^\circ = -110.5 \, \text{kJ/mol} \)
- For \( \text{H}_2\text{O}(g) \): \( \Delta H_f^\circ = -241.8 \, \text{kJ/mol} \)
- For \( \text{H}_2(g) \): \( \Delta H_f^\circ = 0 \, \text{kJ/mol} \) (as it is in its standard state)

2. **Write the Formula for Enthalpy Change**:
The standard enthalpy change (\( \Delta H \)) for the reaction can be calculated using the following formula:
\[ \Delta H = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) \]

3. **Calculate the Enthalpy of Formation for Products**:
- Products: \( \text{CO}(g) \) and \( \text{H}_2\text{O}(g) \)
\[ \Delta H_f^\circ \text{ of products} = \Delta H_f^\circ (\text{CO}) + \Delta H_f^\circ (\text{H}_2\text{O}) = (-110.5) + (-241.8) = -352.3 \, \text{kJ/mol} \]

4. **Calculate the Enthalpy of Formation for Reactants**:
- Reactants: \( \text{CO}_2(g) \) and \( \text{H}_2(g) \)
\[ \Delta H_f^\circ \text{ of reactants} = \Delta H_f^\circ (\text{CO}_2) + \Delta H_f^\circ (\text{H}_2) = (-395.5) + (0) = -395.5 \, \text{kJ/mol} \]

5. **Substitute Values into the Enthalpy Change Formula**:
\[ \Delta H = (-352.3) - (-395.5) = -352.3 + 395.5 = 43.2 \, \text{kJ/mol} \]

6. **Final Result**:
The standard enthalpy change for the reaction is:
\[ \Delta H = +43.2 \, \text{kJ/mol} \]