A mixture of 2 mol of carbon monoxide and 1 mol of oxygen in a closed vessel is ignited to get carbon dioxide , then

Options

A. `DeltaH gt DeltaE`
B. `DeltaH lt DeltaE`
C. `DeltaH = DeltaE`
D. Not definite.

Explanation

To solve the problem, we need to analyze the reaction between carbon monoxide (CO) and oxygen (O2) to form carbon dioxide (CO2) and determine the relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) for the reaction.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between carbon monoxide and oxygen can be represented as:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]

2. **Identify the Moles of Reactants and Products:**
- Moles of carbon monoxide (CO) = 2
- Moles of oxygen (O2) = 1
- Moles of carbon dioxide (CO2) produced = 2

3. **Calculate Δn_g:**
Δn_g is defined as the change in the number of moles of gaseous products minus the change in the number of moles of gaseous reactants.
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- Moles of gaseous products = 2 (from CO2)
- Moles of gaseous reactants = 2 (from CO) + 1 (from O2) = 3
\[ \Delta n_g = 2 - 3 = -1 \]

4. **Use the Relation Between ΔH and ΔE:**
The relationship between enthalpy change (ΔH) and internal energy change (ΔE) is given by:
\[ \Delta H = \Delta E + \Delta n_g \cdot R \cdot T \]
Substituting Δn_g into the equation:
\[ \Delta H = \Delta E + (-1) \cdot R \cdot T \]
This simplifies to:
\[ \Delta H = \Delta E - R \cdot T \]

5. **Rearranging the Equation:**
Rearranging the equation gives:
\[ \Delta E = \Delta H + R \cdot T \]

6. **Conclusion:**
Since R and T are positive values, it follows that:
\[ \Delta E > \Delta H \]
Therefore, the correct answer is that ΔE is greater than ΔH.

### Final Answer:
ΔE > ΔH