250 likes | 498 Views
Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9. The Keq is a constant- a number that does not change. Changing the volume, pressure, or any concentration, does not change the Keq. Only temperature changes the Keq If the [Reactant] is increased the reaction will shift.
E N D
Predicting if a Reaction is in Equilibrium Trial Keq Lesson 9
The Keq is a constant- a number that does not change. Changing the volume, pressure, or any concentration, does not change the Keq. Only temperature changes the Keq If the [Reactant] is increased the reaction will shift
The Keq is a constant- a number that does not change. Changing the volume, pressure, or any concentration, does not change the Keq. Only temperature changes the Keq If the [Reactant] is increased the reaction will shift right The Keq will
The Keq is a constant- a number that does not change. Changing the volume, pressure, or any concentration, does not change the Keq. Only temperature changes the Keq If the [Reactant] is increased the reaction will shift right The Keq will remain constant If the temperature of an endothermic reaction is increased the reaction will
The Keq is a constant- a number that does not change. Changing the volume, pressure, or any concentration, does not change the Keq. Only temperature changes the Keq If the [Reactant] is increased the reaction will shift right The Keq will remain constant If the temperature of an endothermic reaction is increased the reaction will shift right The Keq will
The Keq is a constant- a number that does not change. Changing the volume, pressure, or any concentration, does not change the Keq. Only temperature changes the Keq If the [Reactant] is increased the reaction will shift right The Keq will remain constant If the temperature of an endothermic reaction is increased the reaction will shift right The Keq will increase
1. If 6.00 moles CO2, and 6.00 moles H2are put in a 2.00 L container at 670 oC, calculate all equilibrium concentrations. CO(g) + H2O(g) ⇄ CO2(g) + H2(g)Keq = 9.0 Initial concentrations means ICE! Because we are starting with products it goes left Add on left and subtract on right I 0 0 3.00 M 3.00 M C +x +x -x -x E x x 3.00 - x 3.00 - x Keq = [CO2][H2] = 9.0 [CO][H2O]
Keq = (3 - x)2 = 9 x2 Square root both sides 3 - x = 3 x 1 Cross multiply 3x = 3 - x 4x = 3 [CO] = [H2O] = x = 0.75 M [CO2] = [H2] = 3.00 - 0.75 = 2.25 M
Ktrial How can you tell if a system is in equilibrium or not? Calculate a trial Keq. Put the concentrations into the equilibrium expression and evaluate.
Ktrial How can you tell if a system is in equilibrium or not? Calculate a trial Keq. Put the concentrations into the equilibrium expression and evaluate. If Ktrial = KeqEquilibrium Kt Keq
Ktrial How can you tell if a system is in equilibrium or not? Calculate a trial Keq. Put the concentrations into the equilibrium expression and evaluate. If Ktrial = KeqEquilibrium If Ktrial < KeqNot at Equilibrium Keq Kt
Ktrial How can you tell if a system is in equilibrium or not? Calculate a trial Keq. Put the concentrations into the equilibrium expression and evaluate. If Ktrial = KeqEquilibrium If Ktrial < KeqNot at Equilibrium Shifts Right Keq Kt
Ktrial How can you tell if a system is in equilibrium or not? Calculate a trial Keq. Put the concentrations into the equilibrium expression and evaluate. If Ktrial = KeqEquilibrium If Ktrial < KeqNot at Equilibrium Shifts Right If Ktrial > Keq Keq Kt
Ktrial How can you tell if a system is in equilibrium or not? Calculate a trial Keq. Put the concentrations into the equilibrium expression and evaluate. If Ktrial = KeqEquilibrium If Ktrial < KeqNot at Equilibrium Shifts Right If Ktrial > KeqNot at Equilibrium Shifts Left Keq Kt
The following amounts of gases are placed into a 2.0 L container. Determine if each system is at equilibrium or not. If not, determine the direction that the equilibrium will shift in order to get to equilibrium. 2NH3(g)⇄ N2(g) + 3H2(g) Keq = 10 1. 2.0 moles NH3 2.0 moles N2 2.0 mole H2 Get concentrations. 1.0 M 1.0 M 1.0 M Calculate Kt Kt = [N2][H2]3 = (1)(1)3 = 1 [NH3]2 (1)2 Not in equilibrium Kt < Keq Shifts right!
The following amounts of gases are placed into a 2.0 L container. Determine if each system is at equilibrium or not. If not, determine the direction that the equilibrium will shift in order to get to equilibrium. 2NH3(g)⇄ N2(g) + 3H2(g) Keq = 10 2. 2.0 moles NH3 4.0 moles N2 4.0 mole H2 Get concentrations. 1.0 M 2.0 M 2.0 M Calculate Kt Kt = [N2][H2]3 = (2)(2)3 = 16 [NH3]2 (1)2 Not in equilibrium Kt > Keq Shifts left!
The following amounts of gases are placed into a 2.0 L container. Determine if each system is at equilibrium or not. If not, determine the direction that the equilibrium will shift in order to get to equilibrium. 2NH3(g)⇄ N2(g) + 3H2(g) Keq = 10 3. 2.0 moles NH3 2.5 moles N2 4.0 mole H2 Get concentrations. 1.0 M 1.25 M 2.0 M Calculate Kt Kt = [N2][H2]3 = (1.25)(2)3 = 10 [NH3]2 (1)2 In equilibrium Kt = Keq
4. If 4.00 moles of CO, 4.00 moles H2O, 6.00 moles CO2, and 6.00 moles H2 are placed in a 2.00 L container at 670 oC, Keq = 1.0 CO(g) + H2O(g) ⇄ CO2(g) + H2(g) Is the system at equilibrium? If not, how will it shift in order to get there? Calculate all equilibrium concentrations. Get Molarities 2.00 M 2.00 M 3.00 M 3.00 M Calculate a Kt Kt = (3)(3) = 2.25 (2)(2) Not in equilibrium Shifts left!
Do an ICE chart CO(g) + H2O(g) ⇄ CO2(g) + H2(g) I 2.00 M 2.00 M 3.00 M 3.00 M C +x +x -x -x E 2.00 + x 2.00 + x 3.00 - x 3.00 - x Keq = (3 - x)2 = 1.0 (2 + x)2 Square root 3 - x = 1.0 2 + x 3 - x = 2 + x 1 = 2x x = 0.50 M [CO2] = [H2] = 3.00 - 0.50 = 2.50 M [CO] = [H2O] = 2.00 + 0.50 = 2.50 M
Big Keq Keq = Keq = 10 products reactants
Little Keq- Keq = Keq = 0.1 Note that the keq cannot be a negative number! products reactants
Keq about 1 Keq = Keq = 1 products reactants
Which reaction favours the products the most? Keq = 2.6 x 106 Keq = 3.5 x 102 Which reaction favours the reactants the most? Keq = 2.6 x 10-6 Keq = 3.5 x 10-7
Which reaction favours the products the most? Keq = 2.6 x 106 Keq = 3.5 x 102 Which reaction favours the reactants the most? Keq = 2.6 x 10-6 Keq = 3.5 x 10-7