Question

Equilibrium and Thermodynamics problem:

For the reaction
NH4Cl (s) ----> NH3 (g) + HCl (g)

delta H = 176 kJ and delta G = 91.2 kJ at 298 K. What is the value of delta G at 1000 K?

A. -109 kJ
B. -64 kJ
C. 64 kJ
D. 109 kJ

You don't have to explain it completely, I just want to see someone else's answer. I keep getting -705 kJ.

Answer 1

We are interested in the following expression which you can show:
\[\left( \frac{ \partial G }{ \partial T } \right)_{p}=-S\]
The entropy can be written as:
\[S=\frac{ (H-G) }{ T }\]
We therefor get:
\[\left( \frac{ \partial G }{ \partial T } \right)_{p}=\frac{ (G-H) }{ T }\]
\[\left( \frac{ \partial G }{ \partial T } \right)_{p}-\frac{ G }{ T }=-\frac{ H }{ T }\]
Combine the two terms on the left side:
\[\left( \frac{ \partial }{ \partial T } \frac{ G }{ T } \right)_{p}=\frac{ 1 }{ T } \left( \frac{ \partial G }{ \partial T } \right)_{p}+G \frac{ d }{ dT }\frac{ 1 }{ T }=\frac{ 1 }{ T } \left( \frac{ \partial G }{ \partial T } \right)_{p}-\frac{ G }{ T ^{2} }\]

Answer 2

We therefor get:
\[\left( \frac{ \partial }{ \partial T } \frac{ G }{ T } \right)_{p}=\frac{ 1 }{ T }\left\{ \left( \frac{ \partial G }{ \partial T } \right)_{p}-\frac{ G }{ T } \right\}\]

Answer 3

We use the very first partial differential and substitute:
\[\left( \frac{ \partial }{ \partial T } \frac{ G }{ T } \right)_{p}=\frac{ 1 }{ T }\left\{ -S -\frac{ G }{ T } \right\}=\frac{ 1 }{ T }\left\{ -S-\frac{ H-TS }{ T } \right\}=-\frac{ H }{ T ^{2} }\]
Integreate the expression below using separation of the variables and you have your equation.
\[\left( \frac{ \partial }{ \partial T } \frac{ G }{ T } \right)_{p}=-\frac{ H }{ T ^{2} }\]

Answer 4

As we do not have a heat capacity at constant pressure we approximate and assume that the enthalpy is independent of the change in temperature.

Answer 5

Doing so and using that Delta G is proportional to Delta H we get (tell me if you want to see the integral:
\[\large \Delta H \left( \frac{ 1 }{ T _{2} }-\frac{ 1 }{ T _{1} } \right)=\frac{ \Delta G _{2} }{ T _{2} }-\frac{ \Delta G _{1} }{ T _{1} }\]
Solve for Delta G_2:
\[\large \Delta G _{2}=\Delta H \left( \frac{ T _{2} }{ T _{2} }-\frac{ T _{2} }{ T _{1} } \right)+\frac{ \Delta G _{1} ~ T _{2} }{ T _{1} }=\Delta H \left( 1-\frac{ T _{2} }{ T _{1} } \right)+\frac{ \Delta G _{1} ~ T _{2} }{ T _{1} }\]
For evaluation see attachment.

Answer 6

Wow, this is very in depth, thank you!

Answer 7

No problem. The equation before I solve for Delta G_2 is called the Gibbs–Helmholtz equation. :)

Answer 8

Or perhaps more the partial derivative as you can approximate depending on how accurate you want the answer.