# Why does heat of reaction equal the change in enthalpy for chemical reactions?

Chemists say something like "amount of heat consumed for a chemical reaction equals the change in enthalpy" but I cannot understand why this is the case.

Here is my argument: Since $$H = U +PV$$, we have $$dH = T dS + V dP + \sum_i \mu_i dN_i$$. If we assume that the heat flow is quasistatic so that we can use $$dQ=TdS$$, and assuming that $$P$$ is constant during the reaction so that $$dP=0$$, we have $$dH = dQ + \sum_i \mu_i dN_i$$.

Apparently we have an additional term $$\sum_i \mu_i dN_i$$, so that $$dH \neq dQ$$.

Where am I wrong?

## 1 Answer

For a closed system (no mass transfer into or out of system) at constant pressure, $$\Delta U=Q-P\Delta V$$This equation applies irrespective of whether a chemical reaction is occurring within the system. So, $$\Delta H=\Delta U+P\Delta V=Q$$The heat of reaction is also defined such that T does not change between the initial and final states of the system.

Also, how can it be quasi static if there is a chemical reaction occurring, presumably at finite rate?

• So Dr. Miller, are you saying that the fault in my reasoning is the assumption of quasi-staticity? BTW, thanks for your derivation for $\Delta H = Q$ ! Dec 12, 2020 at 15:47
• I would say yes. It takes some doing to make a reaction proceed quasisstatcly. Google van't Hoff equilibrium box. Dec 12, 2020 at 15:57