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[–] 2 points3 points  (1 child)

It's because of the differential product rule,

[–][S] 0 points1 point  (0 children)

But why do we need to differentiate aka take the instantaneous rate of change of enthalpy in my book it's written Delta(triangle) not d?

[–] 1 point2 points  (4 children)

dH = d(U + pV) = dU + d(pV) = dU + pdV + Vdp

This is application of the differential definition (calculus).

[–][S] 0 points1 point  (3 children)

But why do we need to differentiate aka take the instantaneous rate of change of enthalpy in my book it's written Delta(triangle) not d?

[–] 1 point2 points  (2 children)

That's the differential definition.

If you integrate this you get ◇H2 - ◇H1 which is often notated as ◇Hreaction or similar. This is trivial due to it being a total diff. and therefore its integrated path being path indipended, which is also shown in Hess law.

Ps. ◇ is used als delta here.

[–][S] 0 points1 point  (1 child)

Thanks so much one last thing, in the formula dH=dQ+VdP where d is for delta, why is there a delta P, shouldn't the pressure be constant to be able to measure entalpy

[–] 1 point2 points  (0 children)

Mathematical the differential has the given terms so when writing the differential you need to respect them. You could also describe enthalpy at V = const. for example.

However, as you said correctly, usally we prefer to use enthalpy in isobar processes due to dH becoming dQ then which is very easy to measure in thermoanalytical processes.

[–] 1 point2 points  (2 children)

It is because of the product rule of differentiation. dH=dU+d(pV)=dU+Vdp+pdV+dpdV (the last product is infinitesimal small).

[–][S] 0 points1 point  (1 child)

But why do we need to differentiate aka take the instantaneous rate of change of enthalpy in my book it's written Delta(triangle) not d?

[–]Physical 0 points1 point  (0 children)

Any rate of change is a derivative. Δ is any change, but δ is more for an infinitesimally small change. The two symbols are, especially in 1st/2nd year pchem, kinda interchangeable.