Change in enthalpy in an isolated, isochoric system

Question

Suppose we had an isolated, isovolumetric system in which the reaction

$${N_2 + 2H_2 \to 2NH_3}$$

Takes place. If we had to determine the signs of heat, work, change in internal energy and change in enthalpy, this would be my approach but I don't know if it makes sense.

It is an isolated system, so $Q=W= \Delta U = 0$

$\text{Change in enthalpy}=\Delta U+\Delta (pV) = V\Delta (p)$

Since the number of moles is decreasing, the pressure must decrease as well considering there is no volume change meaning $\Delta H$ is negative.

I also don't know if temperature is constant or if it is changing because I thought enthalpy was a function of temperature as well.

I'm having a hard time grasping how change in internal energy can be zero and how change in enthalpy is negative. I also don't know if my reasoning is correct.

Thank you.

Answer 1

Your analysis, for the most part, is correct. However, you have not taken into account the fact that internal energy and enthalpy are functions not only of temperature but also of chemical composition, which changes as a result of the reaction.

The heat of formation (enthalpy) of ammonia from N2 and H2 at standard temperature and pressure is -46 kJ/mole. This means that, unless approximately this amount of heat is removed, the temperature will rise. In your isochoric system, which is also adiabatic, the change in internal energy will be zero, which means that the approximately 46 kJ/mole will cause the temperature of the product mixture to rise by a corresponding amount. This will allow the internal energy change of the mixture to be zero.

The change in enthalpy of the mixture will not be zero, as you pointed out. But the effect of the change in the number of moles on $\Delta (PV)$ will be somewhat compensated by the increase in temperature. The net effect would have to be worked out, knowing the final state.