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) from the heat of reaction at constant volume (ΔU), we can use the following relationship:

\[ \Delta H = \Delta U + \Delta N_g RT \]

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

### Step 1: Convert the temperature from Celsius to Kelvin
Given temperature is \(17^\circ C\).

\[ T(K) = 17 + 273 = 290 \text{ K} \]

### Step 2: Identify ΔU
The heat of reaction at constant volume is given as:

\[ \Delta U = -67.71 \text{ kcal} \]

### Step 3: Calculate ΔN_g
The reaction is:

\[ CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g) \]

To find ΔN_g, we calculate the change in the number of moles of gas:

- Moles of gaseous products = 1 (from CO2)
- Moles of gaseous reactants = 1 (from CO) + 0.5 (from O2) = 1.5

Thus,

\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 1 - 1.5 = -0.5 \]

### Step 4: Calculate the term ΔN_g RT
Using R = 2 cal/(K·mol) (or 0.002 kcal/(K·mol) for our calculation):

\[ \Delta N_g RT = -0.5 \times 2 \text{ cal/(K·mol)} \times 290 \text{ K} \]

Calculating this:

\[ \Delta N_g RT = -0.5 \times 2 \times 290 = -290 \text{ cal} = -0.29 \text{ kcal} \]

### Step 5: Substitute values into the equation for ΔH
Now we substitute ΔU and ΔN_g RT into the equation for ΔH:

\[ \Delta H = \Delta U + \Delta N_g RT \]

\[ \Delta H = -67.71 \text{ kcal} - 0.29 \text{ kcal} \]

\[ \Delta H = -68.00 \text{ kcal} \]

### Final Answer
The heat of reaction at constant pressure at \(17^\circ C\) is:

\[ \Delta H = -68 \text{ kcal} \]