Question

If Kc = 0.405 at 40.°C and Kc = 0.575 at 90.°C, what is ΔH° for the reaction?
RXN is simply x yields y and reverse

Answer 1

Van Hoff equation mate.

Answer 2

But do I need to rearrange it?

Answer 3

You have to solve for \(\Delta H^{\circ}\) of cause.
But the van't Hoff equation is given by:
\[\Large \frac{ d \ln(K) }{ dT }=\frac{ \Delta H^{\circ} }{ RT^{2} }\]
Solve the differential equation we get
\[\Large \ln(\frac{ K_{2} }{ K_{1} })=\frac{ - \Delta H^{\circ} }{ R } \times \left( \frac{ 1 }{ T_{2} }-\frac{ 1 }{ T_{1} } \right)\]

Solve for standard enthalpy and we are there.

Answer 4

Can you help me solve for h?

Can't you just use algebra instead of calculus?

Answer 5

I don't see what the problem is. The last part is just algebra (solving for \(\Delta H\))?

Answer 6

You don't have to solve the differential equation.. I've already done that for you.

Answer 7

Thanks.

Answer 8

Else let me fast show how I fast figure out how to solve the problem.

We are interesting in some expression with
\[\Large \frac{ \partial K }{ \partial T }\]
We can relate the Gibbs free energy to the equilibrium constant from the known equation
\[\Large \Delta G^{\circ}=-RT \ln(K)\]\[\Large \ln(K)=-\frac{ \Delta G^{\circ} }{ RT }\]
Differentiation of ln K with respect to temperature
\[\Large \frac{ d }{ dT }\ln(K)=-\frac{ 1 }{ R }\frac{ d }{ dT }\left( \frac{ \Delta G^{\circ} }{ T } \right)\]
Using the Gibbs-Helmholtz expression we can evaluate the differential for Gibbs free energy. The Gibbs-Helmholtz expression is given by
\[\Large \left( \frac{ \partial }{ \partial T } \frac{ \Delta G ^{\circ} }{ T } \right)_{p}=-\frac{ \Delta H ^{\circ} }{ T^2 }\]
Substute we get the Van't Hoff equation
\[\Large \Large \frac{ d \ln(K) }{ dT }=\frac{ \Delta H^{\circ} }{ RT^{2} }\]
Now we need solve the equation. We choose to assume that \(\Delta H^{\circ}(T)\) is constant in the temperature interval [\(T_{1},T_{2}\)], we can then take \(\Delta H^{\circ}(T)\) out of the integral and treat it as a constant. Using separation of variables and integration in the respective intervals we get
\[\Large \ln(T_{2})-\ln(T_{1})=\frac{ \Delta H^{\circ} }{ R }\int\limits_{T_{1}}^{T_{2}}T ^{-2} dT\]
Solve the integral we get the solution I wrote
\[\Large \ln(K_{2})-\ln(K_{1}) =\ \frac{ - \Delta H^{\circ} }{ R } \times \left( \frac{ 1 }{ T_{2} }-\frac{ 1 }{ T_{1} } \right)\]
Solve for the standard enthalpy, \(\Delta H^{\circ}\)
\[\Large \Delta H^{\circ}=-R\left( \frac{ \ln(K _{2})-\ln(K_{1}) }{\frac{ 1 }{ T_{2}}-\frac{ 1 }{ T_{1} } } \right)\]

Answer 9

Sorry in the 7th expression I wrote
\[\large \ln(T_{2})-\ln(T_{1})\]
I meant
\[\large \ln(K_{2})-\ln(K_{1})\]
It should however not totally destroy the mathematical context.

Answer 10

If you like to have fun with thermodynamics we can choose to change a bit in our approximations.
In the previous method we chooses to assume that
\[\Large \left( \frac{ \partial \Delta H^{\circ} }{ \partial T } \right)_{p}=\Delta C _{p}^{\circ}=0\]
We can make a more exact calculation and this time assume that the change in standard heat capacity, \( \Delta C_{p}^{\circ}\) remains constant in the temperature interval [\(T_{1},T_{2}\)], so our approximation build on
\[\Large \left( \frac{ \partial ^{2} \Delta H^{\circ} }{ \partial T^2 } \right)_{p}=\left( \frac{ \partial \Delta C _{p}^{\circ} }{ T } \right)_{p}=0\]
Doing this we can solve the Van't Hoff equation to
\[\ln(K_{2})-\ln(K_{1})=-\frac{ \Delta H^{\circ}(T_{1})- \Delta C_{p}^{\circ} T_{1} }{ R }\left( \frac{ 1 }{ T_{2} }-\frac{ 1 }{ T_{1} } \right)+\frac{ 1 }{ R } \Delta C_{p}^{\circ} \ln \left( \frac{ T_{2} }{ T_{1} } \right)\]
This expression is more complex as we now take into account that the enthalpy is temperature dependent.

Answer 11

Correction!
\[\large \left( \frac{ \partial \Delta C_{p}^{\circ} }{ \partial T } \right)_{p}\]
Forgot a partial sign the mathematical expression of the second possible approximation.