The different between `DeltaH and DeltaE` for the reaction `BaCl_(2(aq)) + K_(2)SO_(4(aq)) rarr BaSO_(4(s)) darr +2KCl_((aq))`

Options

A. RT
B. 2RT
C. (1/2)RT
D. zero

Answer

To find the difference between ΔH (change in enthalpy) and ΔE (change in internal energy) for the given reaction:

**Step 1: Write the reaction and identify the phases of each component.**
The reaction is:
\[ \text{BaCl}_{2(aq)} + \text{K}_{2}\text{SO}_{4(aq)} \rightarrow \text{BaSO}_{4(s)} + 2\text{KCl}_{(aq)} \]

In this reaction:
- BaCl₂ is in aqueous phase (aq)
- K₂SO₄ is in aqueous phase (aq)
- BaSO₄ is in solid phase (s)
- KCl is in aqueous phase (aq)

**Step 2: Apply the relationship between ΔH and ΔE.**
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) is given by the equation:
\[ \Delta H = \Delta E + nRT \]
Where:
- n = number of moles of gas produced or consumed
- R = universal gas constant
- T = temperature in Kelvin

**Step 3: Determine the number of moles of gas (n).**
In the given reaction, we need to identify the gaseous products and reactants. Here, both BaCl₂ and K₂SO₄ are in aqueous phase, while BaSO₄ is a solid, and KCl is also in aqueous phase.

Since there are no gaseous products or reactants in this reaction, we have:
\[ n = 0 \]

**Step 4: Substitute n into the equation.**
Substituting n = 0 into the equation gives:
\[ \Delta H - \Delta E = 0 \cdot RT \]
This simplifies to:
\[ \Delta H - \Delta E = 0 \]

**Step 5: Conclude the difference between ΔH and ΔE.**
From the above equation, we can conclude that:
\[ \Delta H = \Delta E \]

Thus, the difference between ΔH and ΔE for the reaction is:
\[ \Delta H - \Delta E = 0 \]

**Final Answer:**
The difference between ΔH and ΔE for the reaction is 0.

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