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 K increases Shifts to reactants 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.

Example − 4.144𝑥105 ∗8.31=3.44𝑥106𝐽/𝑚𝑜𝑙 Slope=-DH Use the graph to answer the following: Is the reaction endothermic or exothermic? Explain. What is DH Slope=-DH y= 4.144x105x+2559 Slope is +, so DH is - Exothermic − 4.144𝑥105 ∗8.31=3.44𝑥106𝐽/𝑚𝑜𝑙

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) Treat as normal equilibrium problem: 56 g of CaO is mixed with 22g of CO2 in a 0.1L container. Find [CO2] and amount of moles of CaCO3 formed. Kc @ some temp= 0.30 Kc = 0.30 = 1 [𝐶 𝑂 2 ] Since Kc only relies on CO2 it is always, 1/[CO2] (as long as there is enough CO2) [CO2]= 1 0.3 =3.33𝑀

Example: (gas is limiting) Treat as normal equilibrium problem: 56 g of CaO is mixed with 22g of CO2 in a 0.10L container. Find [CO2] and amount of moles of CaCO3 formed. 1 mols 0.5 mols Kc @ some temp= 0.30 I 5.00 mols/L Q=0.20 < K shifts right -x +x C -x 5.00 - x E x=1.67 mol/L= Kc = 0.3 = 1 5.00−𝑥 change in mol CO2=change in mol CaCO3 =1.67 mol/L*0.10L=0.167 mol

Example: (solid/liquid is limiting) Treat as normal stoichiometry problem if Q<K, or normal equilibrium if Q>K: 28 g of CaO is mixed with 44g of CO2 in a 0.10L container. Find [CO2] and amount of moles of CaCO3 formed. Use the moles to show us which is limiting 0.50 mols 1 mols Q=0.2 < K wants to shift to right, but can’t. CaO is limiting so after reaction: 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. 0 mols CaO 0.5 mols CaO 0.5 mols CaO 5.0 M CaO Starting moles-moles used up from CaO reacting. Moles calculate from CaO (limiting reagent) The solid limiting reagent goes to zero