Which of the indicated relationship is correct for the following exothermic reaction carried out at constant pressure?
`CO(g) +3H_(2)(g ) rarrCH_(4)(g)+H_(2)O(g)`

Options

A. `DeltaU = DeltaH`
B. `DeltaU gt DeltaH`
C. `w lt 0`
D. `q gt 0`

Explanation

To solve the problem, we need to establish the relationship between the change in internal energy (ΔU) and the change in enthalpy (ΔH) for the given exothermic reaction at constant pressure. The reaction is:

\[ \text{CO(g)} + 3\text{H}_2(g) \rightarrow \text{CH}_4(g) + \text{H}_2\text{O}(g) \]

### Step 1: Identify the reaction components
We have:
- Reactants: 1 mole of CO and 3 moles of H₂
- Products: 1 mole of CH₄ and 1 mole of H₂O

### Step 2: Calculate the change in the number of moles of gas (Δn_g)
The change in the number of moles of gas (Δn_g) is calculated as follows:

\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} \]

- Moles of products = 1 (CH₄) + 1 (H₂O) = 2 moles
- Moles of reactants = 1 (CO) + 3 (H₂) = 4 moles

Thus,

\[ \Delta n_g = 2 - 4 = -2 \]

### Step 3: Use the relationship between ΔH and ΔU
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) at constant pressure is given by:

\[ \Delta H = \Delta U + \Delta n_g \cdot R \cdot T \]

Where:
- R = universal gas constant
- T = temperature in Kelvin

### Step 4: Substitute Δn_g into the equation
Substituting Δn_g into the equation, we have:

\[ \Delta H = \Delta U + (-2) \cdot R \cdot T \]

This simplifies to:

\[ \Delta H = \Delta U - 2RT \]

### Step 5: Analyze the relationship
From the equation \( \Delta H = \Delta U - 2RT \), we can see that:

\[ \Delta H < \Delta U \]

This indicates that the change in enthalpy (ΔH) is less than the change in internal energy (ΔU) for this exothermic reaction at constant pressure.

### Conclusion
The correct relationship for the given exothermic reaction at constant pressure is:

\[ \Delta H < \Delta U \]