# Internal energy change for reactions at constant volume vs constant pressure

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.

## 1 Answer

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.