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A question asks us to derive a relationship between the enthalpy change and the change in internal energy for a reaction. Now, I know the relationship is: $\Delta H = \Delta U + V \Delta p + p \Delta V$ But i thought enthalpy was conveniently defined this way ( not derived) so that enthalpy change equals heat change at constant pressure. So which

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closed as unclear what you're asking by M.A.R. ಠ_ಠ, Klaus-Dieter Warzecha, pH13 - Yet another Philipp, Curt F., hBy2Py Mar 24 '16 at 12:19

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First a brief note on the mathematics of Legendre Transforms

Consider a function in two independent variables: $ f(x,y)$

We right the differential of this function as follows:

$df(x,y) = \Big( \frac{\partial f}{\partial x}\Big)_ydx + \Big( \frac{\partial f}{\partial y}\Big)_xdy$

Now, I shall define $u := \Big( \frac{\partial f}{\partial x}\Big) $ and $w := \Big( \frac{\partial f}{\partial y}\Big)$

so, $df = udx + wdy$ [equation 1]

Here, u and x (and similarly w and y) are called conjugate variables. (i.e $ux$ and $wy$ will have the same units as $f$)

[Think of it this way: In thermodynamics, PV, TS have the same units as H, U,G,A]

Now, our goal is transform our original function $f(x,y)$ to a function $g(x,w)$. We wish to replace y, with it's conjugate variable and the function $f $ and $g$ will have the same units.

so, we take the differential of the product $wy$ : $d(wy) = ydw + wdy$

We subtract this from [equation 1] and get $d(f-wy) = udx - ydw $ and can deduce that our Legendre Transformed function $ g(x,w) := f-wy$

Now, we can easily apply this method to Internal Energy (U) to obtain Enthalpy (H)

We have $U(S,V)$ and $dU(S,V) = \Big( \frac{\partial U}{\partial S}\Big)_VdS + \Big( \frac{\partial U}{\partial V}\Big)_SdV $

Now, here we seek $H(S,P) \equiv g$

a) $f \equiv U$

b) $S \equiv x$ is the variable we hold fixed

c) $V \equiv y$ is the variable we switch

d) $\Big( \frac{\partial U}{\partial S}\Big)_V := T \equiv u $(the conjugate of the variable we are holding in place

e) $\Big( \frac{\partial U}{\partial V}\Big)_S := -P \equiv w$ (the conjugate of the variable we change

$ g =f-wy $

so, $H(S,P) = U-(V)(-P)$

Or, $H(S,P) = U + PV$

From here, you can proceed to write down: $dH = dU + PdV + Vdp$

and for macroscopic changes: $\Delta H = \Delta U + P\Delta V + V\Delta P $

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You can define enthalpy through its difference, so in this sense you are right. But maybe the question that have been asked to you refers to an specific kind of system or specifics restrictions. I am just guessing.

For example, if it is about reaction between ideal gases at constant pressure and temperature, you'll have the general expression

$$ \Delta H = \Delta U + \Delta (pV)= \Delta U + p \Delta V $$

which for this particular case, as $pV= nRT$, so $\Delta(pV)= \Delta(nRT)$, and under this ligatures $p \Delta V = RT \Delta n$, so

$$ \Delta H = \Delta U + RT \Delta n$$

Other way the question makes not sense.

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