[Zopiclone]

Hi! I'm new to the forums and currently reading about Chemical Thermodynamics. So here's what I know:

ΔE = q - w

So for constant volume reactions, no work is done hence:

ΔE = q

But for constant pressure reactions, heat be may released (for exothermic reactions) and work is done hence:

ΔE = q - pΔV

so ΔE ≠ q because some of the heat does work.

So we define enthalpy H = E + PV, so, ΔH = ΔE + PΔV = q - pΔV + pΔV = q

and hence for constant pressure reactions ΔH = q.

My confusion is that the book says that the work term pΔV = (Δn)RT for these processes, but how can we hold the temperature constant since it is obviously changing as a result of the heat of the reaction?

Any help would be greatly appreciated. Thanks!

[mjc123]

But for constant pressure reactions, heat be may released (for exothermic reactions)

Are you assuming no heat is released for constant volume reactions? Why?

how can we hold the temperature constant since it is obviously changing as a result of the heat of the reaction?

Is it? Under isothermal conditions it is not, by definition. It depends how efficient is the heat transfer between system and surroundings. If it is efficient, the heat of reaction is lost to (or taken from) the surroundings and T does not change. In real systems it is never perfectly efficient, and there will be a temporary rise or fall of temperature before thermal equilibrium is reestablished, but the important thing is that the final state of the system has the same T as the initial state. If the system is thermally isolated (adiabatic conditions) then the heat of reaction produces a permanent change of temperature. When we measure a temperature change in order to determine ΔH, we try to make the conditions as adiabatic as we can.