Question

Consider the following reaction

CO2(g)+4HF(g)→CF4(g)+2H2O(g)

(c) How, if at all, would Δ_rU differ if the reaction was to form liquid water instead of gaseous water? Justify your response.

Answer 1

By definition, ΔU represents:

$\Delta U=q+w=q-p\Delta V$

Where q is the heat exchanged in the process and w is the work of expansion of contraction of the system.

In our reaction, the heat is related to the rupture and formation of chemical bonds, which will not change if the product is liquid water, since it is still two H atoms bonded to an O atom. What will differ will be the work, since the change in volume ΔV will not be the same if we obtain liquid water.
Since the volume occupied by a liquid is much lower than the volume occupied by a gas, if we obtain liquid water, the ΔV of the reaction will be lower than if we obtain gaseous water. Looking at the expression of ΔU, this means that, with a decrease in ΔV, we will have an increase in ΔU at constant pressure, due to the change in the work done by/to the system during the reaction.