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

Answer

To solve the problem, we need to analyze the reaction between carbon monoxide (CO) and oxygen (O2) to produce carbon dioxide (CO2). The balanced chemical equation for this reaction is:

\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]

### Step-by-Step Solution:

1. **Identify the Moles of Reactants:**
- We have 2 moles of carbon monoxide (CO).
- We have 1 mole of oxygen (O2).

2. **Write the Balanced Reaction:**
- The balanced equation is:
\[ 2 \text{CO} + \text{O}_2 \rightarrow 2 \text{CO}_2 \]
- This tells us that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.

3. **Determine the Limiting Reactant:**
- According to the balanced equation, 2 moles of CO require 1 mole of O2.
- We have exactly the required amount of O2 (1 mole) for 2 moles of CO.
- Therefore, both reactants will be completely consumed, and the reaction will proceed to completion.

4. **Calculate the Moles of Products:**
- From the balanced equation, 2 moles of CO produce 2 moles of CO2.
- Thus, we will produce 2 moles of carbon dioxide (CO2).

5. **Calculate the Change in Moles (Δn):**
- The change in moles (Δn) is calculated as:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} \]
- Moles of products = 2 moles of CO2.
- Moles of reactants = 2 moles of CO + 1 mole of O2 = 3 moles.
- Therefore,
\[ \Delta n = 2 - 3 = -1 \]

6. **Relate Change in Enthalpy (ΔH) and Change in Internal Energy (ΔE):**
- The relationship between ΔH and ΔE is given by:
\[ \Delta H = \Delta E + \Delta n \cdot R \cdot T \]
- Where R is the universal gas constant and T is the temperature in Kelvin.
- Substituting Δn:
\[ \Delta H = \Delta E - R \cdot T \]

7. **Conclusion:**
- From the equation, we can see that:
\[ \Delta E = \Delta H + R \cdot T \]
- This implies that ΔE is greater than ΔH when Δn is negative.

### Final Answer:
The change in internal energy (ΔE) is greater than the change in enthalpy (ΔH) by the term \( R \cdot T \).