# Change in enthalpy in an isolated, isochoric system

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.

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.
• No. It's not true for chemical reactions. Chemical reactions involve the energetics of unmaking and making of chemical bonds. So, even at constant temperature and pressure, an exothermic reaction requires you to remove heat in order to hold the temperature constant, and an endothermic reaction requires you to add heat in order to hold the temperature constant. This means the $\Delta H$ is negative for an exothermic reaction (at constant temperature) and positive for an endothermic reaction at constant temperature. Feb 12, 2018 at 20:40