0
$\begingroup$

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?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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?

Improve this answer
$\endgroup$
2
  • $\begingroup$ So Dr. Miller, are you saying that the fault in my reasoning is the assumption of quasi-staticity? BTW, thanks for your derivation for $\Delta H = Q$ ! $\endgroup$
    – J. Doe
    Commented Dec 12, 2020 at 15:47
  • $\begingroup$ I would say yes. It takes some doing to make a reaction proceed quasisstatcly. Google van't Hoff equilibrium box. $\endgroup$
    – Chet Miller
    Commented Dec 12, 2020 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.