Problem Set Sulfuryl Chloride Equilibria

Gaseous Equilibrium Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

6/17/2015 Gaseous Equilibria Reading Assignment: Zumdahl Chapter 6.3, This lecture continues the topic of dynamic equilibrium with examples drawn from gaseous systems. Changes to the system can be predicted by Le Châtelier's Principle.

6/17/2015 Gaseous Equilibria Starting a system from reactants Starting a system from products Predicting equivalent starting conditions

6/17/2015 Water-Gas Shift Reaction The water-gas shift reaction is a useful industrial process to generate hydrogen gas. CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) What will a graph of the steam pressure as a function of time look like? Suppose 1 atm of CO and 1 atm of steam are allowed to react in the presence of a catalyst.

6/17/2015 Water-Gas Shift Reaction CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Time Pressure (atm) CO H2O CO2 H2 What will the graph of the other gas pressuresother gas pressures look like?

6/17/2015 Water-Gas Shift Reaction CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Time Pressure (atm) CO H2O CO2 H2

6/17/2015 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) At 605 K, the equilibrium pressures are P CO = 0.67 atm P H2O = 0.67 atm P CO2 = 0.33 atm P H2 = 0.33 atm What is the numeric value of the equilibrium constant for the reaction?numeric value Water-Gas Shift Reaction

6/17/2015 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Q= HO 2 ()( CO ) PP )( () P 2 P H 2 = = 0.25 = K p (0.33) (0.67) Water-Gas Shift Reaction

6/17/2015 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Suppose 1 atm of CO 2 and 1 atm of H 2 are allowed to react in the presence of a catalyst. Water-Gas Shift Reaction What will a graph of the gas pressures look like?

Time Pressure (atm) CO H2O CO2 H2 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Water-Gas Shift Reaction

6/17/ Time Pressure (atm) CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) CO H2O CO2 H2 Water-Gas Shift Reaction

6/17/2015 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Suppose 0.5 atm each of CO, H 2 O, CO 2 and H 2 are allowed to react in the presence of a catalyst. Water-Gas Shift Reaction What will a graph of the gas pressures look like?

Time Pressure (atm) CO H2O CO2 H2 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Water-Gas Shift Reaction

6/17/2015 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Water-Gas Shift Reaction

6/17/2015 CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Observations Three different starting combinations of reactants and products give the same final results. Equilibrium can be approached from reactants, products or a combination of both. CO 2 (g) + H 2 (g) CO(g) + H 2 O(g) Water-Gas Shift Reaction

6/17/2015 Sulfur Dioxide Oxidation Sulfur dioxide gas reacts with oxygen to give sulfur trioxide gas. The equilibrium constant for this reaction is 3.46 atm -1. If P SO2 is 4.00 atm and P O2 = 3.00 atm initially, what will be the total pressure of the system at equilibrium?

6/17/2015 Sulfur Dioxide Oxidation Process Write a balanced chemical equation. Write the mass-action expression. Set up an “accounting system” that allows you to determine the pressures of the reactants and products as the system attains equilibrium. Determine the final pressure of the system at equilibrium.

6/17/2015 Sulfur Dioxide Oxidation 2 SO 3 (g)2 SO 2 (g) + O 2 (g) () P 2 SO 3 ()() PP 2 2 O 2 K== 3.46 atm -1

6/17/2015 Sulfur Dioxide Oxidation 2 SO 3 (g)2 SO 2 (g) + O 2 (g) 4.00 atm3.00 atm0 atm this has to increase this has to decrease this has to decrease -x-2x+2x

6/17/2015 ICE Table 2 SO 3 (g)2 SO 2 (g) + O 2 (g) SO 2 O2O2 SO 3 Initial Change Equilibrium x-2x+2x x3.00-x2x Substitute into the mass-action expression and solve

6/17/2015 Sulfur Dioxide Oxidation () P 2 SO 3 ()() PP 2 2 O 2 K== 3.46 atm -1 2 SO 3 (g)2 SO 2 (g) + O 2 (g) K = (2x) 2 (4.00-2x ) 2 (3.00-x) = 3.46 atm -1

6/17/2015 Sulfur Dioxide Oxidation K = (2x) 2 (4.00-2x ) 2 (3.00-x) = 3.46 atm -1 solve(((2*x)^2)/(((4-2*x)^2)*(3-x))=3.46,x); , I, I 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 ICE Table SO 2 O2O2 SO 3 Initial Change Equilibrium x-x+2x x3.00-x2x *1.40 = 1.20 atm = 1.60 atm 2*1.40 = 2.80 atm 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Sulfur Dioxide Oxidation K = (2.80 atm) 2 (1.20 atm ) 2 (1.60 atm) = 3.46 atm -1 = 3.40 atm -1 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Sulfur Dioxide Oxidation Total Pressure = P SO2 + P O2 + P SO3 =( ) atm = 5.60 atm 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Equivalent Starting Conditions Example: In the previous example, how much sulfur trioxide gas could have been placed in an evacuated container to end up with the same equilibrium conditions?

6/17/2015 Sulfur Dioxide Oxidation 4.00 atm3.00 atm0 atm 2 SO 3 (g)2 SO 2 (g) + O 2 (g) Suppose 2 atm of SO 2 reacts, what will be the resulting pressures?

6/17/2015 Sulfur Dioxide Oxidation 4.00 atm3.00 atm0 atm 2.00 atm 0 atm1.00 atm4.00 atm Equilibrium is shifted as far left as possible Equilibrium is shifted as far right as possible 2 SO 3 (g)2 SO 2 (g) + O 2 (g) Suppose 2 atm of SO 2 reacts, what will be the resulting pressures? Suppose it shifts as far to the right as possible, what will be the resulting pressures?

6/17/2015 ICE Table SO 2 O2O2 SO 3 Initial Change Equilibrium y+2y-2y 2y1.00+y4.00-2y 2*0.60 = 1.20 atm = 1.60 atm *0.60 = 2.80 atm 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Perturbing a System at Equilibrium Addition or removal of reactant Addition or removal of product Addition of a non-reacting component with no change in volume Volume change Temperature change Le Châtelier's Principle

6/17/2015 Le Châtelier's Principle 2 SO 3 (g)2 SO 2 (g) + O 2 (g) What will be the effect on the reactants and products as each of the following changes are made? Additional SO 2 is added to the system. SO 2 O2O2 SO 3 initially up then down downup

6/17/2015 Le Châtelier's Principle What will be the effect on the reactants and products as each of the following changes are made? O 2 is removed from the system. SO 2 O2O2 SO 3 initially down then up up down 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Le Châtelier's Principle What will be the effect on the reactants and products as each of the following changes are made? Additional SO 3 is added to the system. SO 2 O2O2 SO 3 initially up then down up 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Le Châtelier's Principle What will be the effect on the reactants and products as each of the following changes are made? Inert N 2 is added to the system with no change in volume or temperature. SO 2 O2O2 SO 3 no change no change no change 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Le Châtelier's Principle What will be the effect on the reactants and products as each of the following changes are made? The volume of the system is increased with no change in temperature. SO 2 O2O2 SO 3 initially down then up initially down then up initially down then down 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Changing the Volume of a Gaseous Equilibrium System high pressure small volumes 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Changing the Volume of a Gaseous Equilibrium System less at larger volumes more at larger volumes 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Changing the Volume of a Gaseous Equilibrium System low pressure large volumes 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

Pressure Effects Nitrogen Dioxide-Dinitrogen Tetroxide System

Nitrogen Dioxide - Dinitrogen Tetroxide NO 2 (g)+ NO 2 (g)N 2 O 4 (g) N O O N O O N O O N O O

Nitrogen Dioxide - Dinitrogen Tetroxide 0 sec1 sec10 sec NO 2 (g)+ NO 2 (g)N 2 O 4 (g)

[reactants] [products] Nitrogen Dioxide - Dinitrogen Tetroxide Mass-Action Expression Q = Why does the gas initially get darker and then lighten? P NO 2 ( )2)2 P N2O4N2O4 NO 2 (g)+ NO 2 (g)N 2 O 4 (g)

6/17/2015 Gas Pressure Analysis Volume L P SO2 atm P O2 atm P SO3 atm P total atm K p % Reaction /5.72 = 65.0% /0.572 = 43.0% / = 21.9% Expanding the volume of the system causes the partial pressures of all the gases to decrease. 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Gas Pressure Analysis Volume L P SO2 atm P O2 atm P SO3 atm P total atm K p % Reaction /5.72 = 65.0% /0.572 = 43.0% / = 21.9% As the volume increases, the percentage reaction shifts toward the side with more total moles of gas. 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Gas Pressure Analysis Volume L P SO2 atm P O2 atm P SO3 atm P total atm K p % Reaction /5.72 = 65.0% /0.572 = 43.0% / = 21.9% If the equilibrium is shifted all the way RIGHT or LEFT what would be the initial starting conditions? 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Equivalent Starting Conditions This system could have been prepared by initially adding 5.72 atm of SO 3 gas to the evacuated 1.00 L cylinder, or by adding 5.72 atm of SO 2 gas and 2.86 atm of O 2 gas to the evacuated 1.00 L cylinder, or several other possibilities.

6/17/2015 Gas Pressure Analysis Volume L P SO2 atm P O2 atm P SO3 atm P total atm K p % Reaction /5.72 = 65.0% /0.572 = 43.0% / = 21.9% The stoichiometric relationships remain true: SO 2 and O 2 ratio 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Moles of Gas Analysis The stoichiometric relationships remain true: total moles of sulfur containing compounds is constant. Volume L Total moles Moles SO Moles O Moles SO SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Moles of Gas Analysis As the volume expands, the reactions shifts left towards the side with more moles of gas. Volume L Total moles Moles SO Moles O Moles SO SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Mole Fraction Analysis Volume L /6.72 = SO 2  O2O2  SO 3  As the volume expands, the reactions shifts left. 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015 Mole Fraction Analysis Volume L /6.72 = SO 2  O2O2  SO 3  The stoichiometric relationships remain true: SO 2 and O 2 ratio 2 SO 3 (g)2 SO 2 (g) + O 2 (g)

6/17/2015