Internal energy change for reactions at constant volume vs constant pressure

Question

The reaction of cyanamide, $\ce{NH2CN(s)},$ with dioxygen was carried out in a bomb calorimeter, and $∆U$ was found to be $\pu{–742.7 kJ mol-1}$ at $\pu{298 K}.$ Calculate enthalpy change for the reaction at $\pu{298 K}.$

$$\ce{NH2CN(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)}$$

$ΔH$ could be found from

$$ΔH = ΔU + ΔnRT$$

I want to know whether $ΔU$ found for constant volume conditions (bomb calorimeter) can be used in constant pressure conditions (while calculating $ΔH$ pressure is constant). In other words, is $ΔU$ same for both constant pressure and constant volume conditions, and why?

I searched the entire net and several standard physical chemistry texts but couldn't find any explanation.

Answer 1

The key point to keep in mind in this type of problem is that the energy of an ideal gas is only a function of temperature, not of volume or pressure$^\ast$. We also assume that the dependence of the energy of the condensed phases on p and V is negligible. Therefore the final pressure or volume is not going to affect $\Delta U$ for the reaction, provided n and T are constant. Since here the reaction refers to a conversion of a stoichiometric amount of reactants into products at the specified T, we therefore do not expect much of a change in $\Delta U$ with change in p or V.

$^\ast$Note that the total differential for U can be written as $$\begin{align} dU &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV + C_VdT \end{align}$$ Therefore at constant T, $$\begin{align} dU = &= \left\{ T \left( \frac{\partial p}{\partial T} \right)_V -p\right\} dV \end{align}$$ The term in parentheses is zero for an ideal gas.