Heat of reaction for, `CO(g)+1//2O_(2)(g)rarr CO_(2)(g)` at constant V is -67.71 K cal at `17^(@)C` . The heat of reaction at constant P at `17^(@)C` is :-

Options

A. `-68.0` K cal
B. `+68.0` K cal
C. `-67.4` K cal
D. None

Answer

To find the heat of reaction at constant pressure (ΔH) given the heat of reaction at constant volume (ΔU), we can use the following relationship:

\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]

Where:
- ΔH = heat of reaction at constant pressure
- ΔU = heat of reaction at constant volume
- ΔN = change in the number of moles of gas
- R = universal gas constant (approximately 2 kcal/(kmol·K) or 0.001987 kcal/(mol·K))
- T = temperature in Kelvin

### Step-by-Step Solution:

1. **Identify Given Values:**
- ΔU = -67.71 kcal (heat of reaction at constant volume)
- Temperature (T) = 17°C = 17 + 273 = 290 K
- R = 2 kcal/(kmol·K) (using this value for calculations)

2. **Calculate ΔN (Change in Moles of Gas):**
- The reaction is:
\[ CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g) \]
- Moles of reactants = 1 (for CO) + 0.5 (for O2) = 1.5 moles
- Moles of products = 1 (for CO2)
- Therefore,
\[ \Delta N = \text{moles of products} - \text{moles of reactants} = 1 - 1.5 = -0.5 \]

3. **Substitute Values into the Equation:**
\[ \Delta H = \Delta U + \Delta N \cdot R \cdot T \]
\[ \Delta H = -67.71 \text{ kcal} + (-0.5) \cdot (2 \text{ kcal/(kmol·K)}) \cdot (290 \text{ K}) \]

4. **Calculate the Second Term:**
\[ -0.5 \cdot 2 \cdot 290 = -290 \text{ kcal} \]

5. **Final Calculation:**
\[ \Delta H = -67.71 \text{ kcal} - 290 \text{ kcal} \]
\[ \Delta H = -67.71 - 0.58 = -68.0 \text{ kcal} \]

### Conclusion:
The heat of reaction at constant pressure at 17°C is:
\[ \Delta H = -68.0 \text{ kcal} \]