Assume each reaction is carried in open container. For which
reaction `DeltaHgtDeltaE `?

Options

A. `H_(2(g))+I_(2(g))to2HI_((g))`
B. `N_(2(g))+3H_(2(g))to2NH_(3(g))`
C. `PCl_(5(g))toPCl_(3(g))+Cl_(2(g))`
D. `SO_(2(g))+(1)/(2)O_(2(g))toSO_(3(g))`

Explanation

To determine for which reaction the change in enthalpy (ΔH) is greater than the change in internal energy (ΔE) when the reactions are carried out in an open container, we can use the relationship between ΔH and ΔE:

### Step-by-Step Solution:

1. **Understand the relationship between ΔH and ΔE**:
The relationship is given by the equation:
\[ \Delta H = \Delta E + \Delta N_G RT \]
where ΔN_G is the change in the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

2. **Identify the condition for ΔH > ΔE**:
Rearranging the equation, we find:
\[ \Delta H - \Delta E = \Delta N_G RT \]
For ΔH to be greater than ΔE, the term ΔN_G must be greater than 0:
\[ \Delta N_G > 0 \]

3. **Define ΔN_G**:
ΔN_G is defined as the difference between the moles of gaseous products and the moles of gaseous reactants:
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

4. **Analyze the reactions**:
For each reaction, calculate ΔN_G:
- **Reaction 1**: If the reaction has 2 moles of gas products and 1 mole of gas reactants:
\[ \Delta N_G = 2 - 1 = 1 \quad (\text{ΔN_G > 0}) \]
- **Reaction 2**: If the reaction has 2 moles of gas products and 3 moles of gas reactants:
\[ \Delta N_G = 2 - 3 = -1 \quad (\text{ΔN_G < 0}) \]
- **Reaction 3**: If the reaction has 2 moles of gas products and 1 mole of gas reactants:
\[ \Delta N_G = 2 - 1 = 1 \quad (\text{ΔN_G > 0}) \]
- **Reaction 4**: If the reaction has 1 mole of gas products and 1 mole of gas reactants:
\[ \Delta N_G = 1 - 1 = 0 \quad (\text{ΔN_G = 0}) \]

5. **Conclusion**:
From the analysis, we find that ΔN_G is greater than 0 for certain reactions. Therefore, for those reactions where ΔN_G > 0, we can conclude that ΔH > ΔE.