# Overall enthalpy balance on a mixing process

Let's say I have a mixing process involving $$\ce{A}$$ and $$\ce{B}$$ at temperatures $$T_\ce{A}$$ and $$T_\ce{B}$$, respectively. The $$\ce{AB}$$ mix is at temperature $$T_p$$. Our $$ΔH_\mathrm{mix}$$ is defined at $$\pu{25 °C}$$ (i.e., our reactants/products are not at the reference temperature) and we know its value. What would be the overall balance?

It is my understanding that we would have 3 steps:

1. Use the amounts and heat capacities of $$\ce{A}$$ and $$\ce{B}$$ and find $$ΔH$$ (open system) from their specified temperatures to the reference temperature $$\pu{25 °C}$$. This is $$ΔH_1.$$
2. Use the amount and heat capacity of $$\ce{AB}$$ mix to find $$ΔH$$ from the reference temperature to $$T_p$$. This is $$ΔH_2.$$
3. Finally, $$ΔH = ΔH_2 + ΔH_\mathrm{mix} + ΔH_1.$$

That is to say, the overall $$ΔH$$ is the sum of all the steps required to get the initial state(s) to the final state.

I have been told by my lecturer, however, that it should be

$$ΔH = ΔH_2 - ΔH_\mathrm{mix} - ΔH_1$$

Why are $$ΔH_\mathrm{mix}$$ and $$ΔH_1$$ subtracted?

• I think this is just a definition issue. Look closely at how your lecturer defined the $\Delta H$ for each step. Is it for the reaction in the same direction that you used? Using the opposite sign for the last step is one way to emphasize that the $T_p$ is dependent on how much heat was released or absorbed in the first two steps, because that is the amount of heat that must be absorbed or released in the last step, ie $\Delta H_2$ is equal to $-(\Delta H_1 + \Delta H_{mix})$ using your sign convention. (I'm assuming that the system is insulated.) – Andrew Mar 17 '19 at 14:00

Let’s use a trivial case to test the two possible results. If we set $$T_\ce{A}$$, $$T_\ce{B}$$, and $$T_p$$ all to room temperature, the first and third term would be zero (sign does not matter), and you should simply get the enthalpy of mixing as a result. The result you arrived at does that, but the answer you were given changes the sign.