A mixture of 2 moles of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. If `Delta H` is the enthalpy change and `Delta E` is the change in internal energy, then :-

Options

A. `Delta H gt Delta E`
B. `Delta H lt Delta E`
C. `Delta H = Delta E`
D. Not definite

Explanation

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

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between carbon monoxide and oxygen can be represented as:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
Here, we have 2 moles of CO reacting with 1 mole of O2 to produce 2 moles of CO2.

2. **Identify the Number of Moles:**
- Reactants:
- Carbon monoxide (CO): 2 moles
- Oxygen (O2): 1 mole
- Products:
- Carbon dioxide (CO2): 2 moles

3. **Calculate ΔNG (Change in the Number of Moles of Gas):**
ΔNG is calculated as:
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- Moles of gaseous products (CO2): 2 moles
- Moles of gaseous reactants (CO + O2): 2 + 1 = 3 moles
\[ \Delta N_G = 2 - 3 = -1 \]

4. **Use the Relationship Between ΔH and ΔE:**
The relationship between enthalpy change (ΔH) and internal energy change (ΔE) is given by:
\[ \Delta H = \Delta E + \Delta N_G RT \]
Substituting the value of ΔNG:
\[ \Delta H = \Delta E + (-1)RT \]
This simplifies to:
\[ \Delta H = \Delta E - RT \]

5. **Analyze the Relationship:**
From the equation ΔH = ΔE - RT, we can see that ΔH is less than ΔE because we are subtracting a positive term (RT) from ΔE. Therefore:
\[ \Delta H < \Delta E \]

### Conclusion:
The correct relationship is that the enthalpy change (ΔH) is less than the change in internal energy (ΔE).

### Final Answer:
\[ \Delta H < \Delta E \]