The difference between heat of reaction at constant pressure `(DeltaH)` and at constant volume (`DeltaE`) for the reaction :
`C_((s))+(1)/(2)O_(2(g)) to CO_((g))` is :-
(Assume R = 0.002 Kcal ` K^(-1) "mol"^(-1)` and temperature =TK)

Options

A. T cal
B. `-T " Cal " `
C. `-(T)/(2)` cal
D. `(T)/(2)` cal

Explanation

To find the difference between the heat of reaction at constant pressure \((\Delta H)\) and at constant volume \((\Delta E)\) for the reaction:

\[ C_{(s)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)} \]

we can use the relationship:

\[ \Delta H = \Delta E + \Delta N_g R T \]

Where:
- \(\Delta N_g\) = change in the number of moles of gas
- \(R\) = gas constant
- \(T\) = temperature in Kelvin

### Step 1: Calculate \(\Delta N_g\)

1. **Identify gaseous moles in products and reactants**:
- Products: \(CO_{(g)}\) → 1 mole of gas
- Reactants: \(\frac{1}{2} O_{2(g)}\) → 0.5 moles of gas

2. **Calculate \(\Delta N_g\)**:
\[ \Delta N_g = \text{(moles of gas in products)} - \text{(moles of gas in reactants)} = 1 - 0.5 = 0.5 \]

### Step 2: Substitute values into the equation

1. **Substitute \(\Delta N_g\) into the equation**:
\[ \Delta H - \Delta E = \Delta N_g R T \]
\[ \Delta H - \Delta E = 0.5 \cdot R \cdot T \]

2. **Substitute the value of \(R\)**:
Given \(R = 0.002 \, \text{kcal} \, K^{-1} \, mol^{-1}\):
\[ \Delta H - \Delta E = 0.5 \cdot 0.002 \cdot T \]

### Step 3: Simplify the equation

1. **Calculate the difference**:
\[ \Delta H - \Delta E = 0.001 \cdot T \]

### Final Result

The difference between the heat of reaction at constant pressure and constant volume for the given reaction is:

\[ \Delta H - \Delta E = 0.001 T \, \text{kcal} \]