[Foobarz]

So, under constant volume heat is equal to change in energy 1. (delta E = q_v)
And under constant pressure heat is equal to change in enthalpy 2. (delta H = q_p)

And derived from definition of enthalpy (I found all these equations in Chang 8th ed.)
3. H = E + PV
4. delta H =  delta E + delta (PV)
5. delta H - delta (PV) = delta E
6. (if constant pressure) delta H - PdeltaV = delta E
So. . .
Does that mean under constant volume. . .
delta H = delta E ?

But. . .
I think that using the definition of enthalpy,
7. (if constant volume) delta H - VdeltaP = delta E     
so under constant volume (use equation 1) delta H - VdeltaP = q ?

[Foobarz]

An append:

In Chang it says that "because reaction sin a bomb calorimeter occur under constant-volume rather than constant-pressure conditions, the heat changes do not correspond to the enthalpy change delta H" Is my last equation in the previous post needed to correct the q_v into delta H?

[Foobarz]

*Ignore me, I am impatient*

[juanrga]

So, under constant volume heat is equal to change in energy 1. (delta E = q_v)
And under constant pressure heat is equal to change in enthalpy 2. (delta H = q_p)

And derived from definition of enthalpy (I found all these equations in Chang 8th ed.)
3. H = E + PV
4. delta H =  delta E + delta (PV)
5. delta H - delta (PV) = delta E
6. (if constant pressure) delta H - PdeltaV = delta E
So. . .
Does that mean under constant volume. . .
delta H = delta E ?

But. . .
I think that using the definition of enthalpy,
7. (if constant volume) delta H - VdeltaP = delta E     
so under constant volume (use equation 1) delta H - VdeltaP = q ?

H == E + PV

:delta: H =  :delta: E +  :delta: (PV)

If both P and V are constant

:delta: H =  :delta: E

as you say.

If only P is constant

:delta: H =  :delta: E +  P :delta: V

if only V is constant

:delta: H =  :delta: E +  V :delta: P

and using 1 (closed system)

:delta: H =  Q_v +  V :delta: P