Van’t Hoff Equation

Van’t Hoff Equation: Graphs

Experimentally you can use this to determine the DH of reaction.

Van’t Hoff Equation and Le Chatelier’s

As t increases which way does the reaction shift and what happens to K for an endothermic reaction?

As t increases, which way does the reaction shift and what happens to K for an exothermic reaction?

Why are the slopes different signs for endo vs exo thermic?

Shifts to products

Shifts to reactants

K increases

K decreases

Because the sign of the enthalpy is different, changing temperature has the opposite effect on an endothermic reaction as an exothermic reaction, therefore the sign of

the slope will also be opposite.

ExampleUse the graph to answer the following:

Is the reaction endothermic or exothermic? Explain.

What is DH

Slope=-DH

− (4 .144 𝑥 105 )∗8.31=3.44 𝑥106 𝐽 /𝑚𝑜𝑙

y= 4.144x105x+2559

Slope is +, so DH is -

Exothermic

Limiting Regents and Equilibrium.

When do you need to consider this?

Heterogeneous equilibrium!

The amount of CaO and CaCO3 doesn’t matter, so long as

there is enough of each that there is leftover at equilibrium.

Concentration is effectively constant although the mass

changes.

Example: (gas is limiting)

56 g of CaO is mixed with 22g of CO2 in a 0.1L container. Find [CO2] and amount of moles of CaCO3 formed.

Treat as normal equilibrium problem:

Kc = 0.30 =

[CO2]=

Since Kc only relies on CO2 it is always, 1/[CO2] (as long as there is enough CO2)

Kc @ some temp= 0.30

Example: (gas is limiting)

56 g of CaO is mixed with 22g of CO2 in a 0.10L container. Find [CO2] and amount of moles of CaCO3 formed.

Kc @ some temp= 0.30

1 mols

5.00 mols/L

Treat as normal equilibrium problem:

I

C

E

Q=0.20 < K shifts right-x +x-x

0.5 mols

5.00 - x

Kc = 0.3 =x=1.67 mol/L=

change in mol CO2=change in mol CaCO3

=1.67 mol/L*0.10L=0.167 mol

Example: (solid/liquid is limiting)

28 g of CaO is mixed with 44g of CO2 in a 0.10L container. Find [CO2] and amount of moles of CaCO3 formed. 0.50 mols

Treat as normal stoichiometry problem if Q<K, or normal equilibrium if Q>K:

1 mols

CaO is limiting so after reaction:

0 mols CaO 0.5 mols CaO 0.5 mols CaO

Q=0.2 < K wants to shift to right, but can’t.

5.0 M CaO

Use the moles to show us which is limiting

Find Q so that we know if the

reaction will shift. If it is <K it

wants to shift to the right but

can’t because there is no CaO to

react with. Its stuck and stays

where it is.

The solid limiting reagent goes to zero

Starting moles-moles

used up from CaO

reacting.

Moles calculate from

CaO (limiting reagent)