Question

which of the following relationship is correct for a reaction involving both reactants and products are either in solid or liquid state?

Answer 1

ΔH = ΔU

Answer 2

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

$\Delta H = \Delta U + P \Delta V$

where $P$ is the constant pressure and $\Delta V$ is the change in volume during the reaction.

The question states that all reactants and products are either in the solid or liquid state.

For reactions involving only solids and liquids:

1. The volume change ($\Delta V$) during the reaction is typically very small and can be considered negligible. This is because solids and liquids are nearly incompressible, and their volumes do not change significantly upon chemical transformation.
2. Consequently, the term $P \Delta V$ (work done by pressure-volume change) is also very small, almost zero, compared to the magnitudes of $\Delta H$ and $\Delta U$.

Therefore, if $\Delta V \approx 0$, then $P \Delta V \approx 0$.

Substituting this into the general equation:

$\Delta H = \Delta U + 0$ $\Delta H \approx \Delta U$

Thus, for reactions involving only solids and liquids, the enthalpy change is approximately equal to the internal energy change.

This can also be understood in the context of the more general relationship for reactions involving gases: $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants. If there are no gaseous reactants or products, then $\Delta n_g = 0$, which also leads to $\Delta H = \Delta U$.