Question

the difference between delta H and delta E in BaCl2+K2SO4->BaSO4+2KCl

Answer 1

0

Answer 2

The relationship between the change in enthalpy ($\Delta H$) and the change in internal energy ($\Delta E$) for a chemical reaction is given by:

$\Delta H = \Delta E + \Delta (PV)$

For reactions involving gases at constant temperature, this equation can be written as:

$\Delta H = \Delta E + \Delta n_g RT$

where:

• $\Delta n_g$ is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.
• $R$ is the ideal gas constant.
• $T$ is the absolute temperature.

Let's analyze the given reaction:

$\text{BaCl}_2 + \text{K}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{KCl}$

To determine $\Delta n_g$, we need to specify the physical states of the reactants and products. This is a typical precipitation reaction occurring in an aqueous solution. So, the balanced reaction with states is:

$\text{BaCl}_2(aq) + \text{K}_2\text{SO}_4(aq) \rightarrow \text{BaSO}_4(s) + 2\text{KCl}(aq)$

Now, let's count the number of moles of gaseous species:

• Moles of gaseous reactants = 0 (since $\text{BaCl}_2$ and $\text{K}_2\text{SO}_4$ are in aqueous solution)
• Moles of gaseous products = 0 (since $\text{BaSO}_4$ is a solid precipitate and $\text{KCl}$ is in aqueous solution)

Therefore, the change in the number of moles of gaseous substances ($\Delta n_g$) is:

$\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$

$\Delta n_g = 0 - 0 = 0$

Substitute $\Delta n_g = 0$ into the relationship between $\Delta H$ and $\Delta E$:

$\Delta H = \Delta E + (0)RT$

$\Delta H = \Delta E$

The question asks for the difference between $\Delta H$ and $\Delta E$, which is $\Delta H - \Delta E$.

From the above equation, we can write:

$\Delta H - \Delta E = 0$

The difference between $\Delta H$ and $\Delta E$ for this reaction is 0.