Molar latent heat of phosphine

[patrickmoloney]

Homework Statement

[/B]
Phospine exist in three forms. known as the [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] forms. The [itex]\alpha[/itex] and [itex]\beta[/itex] forms are in equilibrium with each other at [itex]49.43 \, K[/itex], and the [itex]\alpha[/itex] and [itex]\gamma[/itex] forms are in equilibrium at [itex]30.29 \, K[/itex]. Obtain the molar heat of transformation for the [itex]\gamma[/itex] form changing to the [itex]\alpha[/itex] form at [itex]30.29 \, K[/itex] from the following data:

(i) The entropy of the [itex]\alpha[/itex] at [itex]49.43 \, K[/itex] is [itex]34.03 \, \text{J/mol K}[/itex].
(ii) The change in entropy heating the [itex]\gamma[/itex] from [itex]0\, K[/itex] to [itex]30.29 \, K[/itex] is [itex]11.22 \, \text{J/mol K}[/itex].
(iii)The change in entropy in heating the [itex]\alpha[/itex] form from [itex]30.29 \, K[/itex] to [itex]49.43 \, K[/itex] is [itex]20.10 \, \text{J/mol K}[/itex]

Homework Equations

The relationship between latent heat [itex]Q_T[/itex] and molar entropy [itex]\Delta S[/itex] is

[tex]Q_T = \Delta S \cdot T[/tex]

[itex][/itex]
The total entropy is calculated by adding the individual entropies of the system.

The Attempt at a Solution

Using the data given we know that [itex]S_{\alpha} = 34.03\, \text{J/mol K}[/itex] at [itex]49.43\, K[/itex]
We also know that the entropy change [itex]\Delta S_{{\alpha}_2}= 20.10\, \text{J/mol K} [/itex] from [itex]30.29 \, K \to 49.43 \, K[/itex]

Hence,

[tex]\Delta S_{\alpha_{1}}= S_{alpha}- \Delta S_{\alpha_{2}} = 34.03 - 20.10 = 13.93 \, \text{J/mol K} [/tex]

is the entropy change in heating the [itex]\alpha[/itex] form from [itex]0 \, K \to 30.29\, K[/itex]

We know the entropy change for [itex]\gamma[/itex] is [itex]\Delta S_{\gamma}= 11.22[/itex]

There for the total entropy [itex]\Delta S_{tot}[/itex] is given by

[tex]\Delta S_{tot} = \Delta S_{alpha_{1}} + \Delta S_{\gamma}= 13.93 + 11.22 = \, 25.15 \text{J/mol K}[/tex]

Our latent heat is

[tex]Q = \Delta S_{tot}\cdot T = (25.15\, \text{J/mol K})(30.29 \, K)= 761.79 \, \text{J/mol}[/tex]is this a correct way to tackle this problem. I struggled for a while cause it's a bit of a weird problem but is it really that straight forward or am I missing something. As they say if physics seems easy - you're doing it wrong.

 

[mjc123]

The total entropy is calculated by adding the individual entropies of the system.

No. You want the change in entropy on going from γ to α. That is Sα - Sγ.