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 CO 2, and 6.00 moles H 2 are put in a 2.00 L container at 670 o C, calculate all equilibrium concentrations. CO (g) + H 2 O (g) ⇄ CO 2(g) + H 2(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 M3.00 M C+x+x-x-x Exx x x Keq=[CO 2 ][H 2 ]=9.0 [CO][H 2 O]

Keq=(3 - x) 2 =9 x 2 Square root both sides 3 - x =3 x1 Cross multiply 3x=3 - x 4x=3 [CO]= [H 2 O] =x=0.75 M [CO 2 ]= [H 2 ] = = 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 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 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 KeqKt

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 KeqKt

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. 2NH 3(g) ⇄ N 2(g) + 3H 2(g) Keq = moles NH moles N mole H 2 Get concentrations. 1.0 M1.0 M1.0 M Calculate Kt Kt=[N 2 ][H 2 ] 3 =(1)(1) 3 = 1 [NH 3 ] 2 (1) 2 Not in equilibriumKt < 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. 2NH 3(g) ⇄ N 2(g) + 3H 2(g) Keq = moles NH moles N mole H 2 Get concentrations. 1.0 M2.0 M2.0 M Calculate Kt Kt=[N 2 ][H 2 ] 3 =(2)(2) 3 = 16 [NH 3 ] 2 (1) 2 Not in equilibriumKt > 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. 2NH 3(g) ⇄ N 2(g) + 3H 2(g) Keq = moles NH moles N mole H 2 Get concentrations. 1.0 M1.25 M2.0 M Calculate Kt Kt=[N 2 ][H 2 ] 3 =(1.25)(2) 3 = 10 [NH 3 ] 2 (1) 2 In equilibriumKt = Keq

4.If 4.00 moles of CO, 4.00 moles H 2 O, 6.00 moles CO 2, and 6.00 moles H 2 are placed in a 2.00 L container at 670 o C, Keq = 1.0 CO (g) + H 2 O (g) ⇄ CO 2(g) + H 2(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 M2.00 M3.00 M3.00 M Calculate a Kt Kt=(3)(3)=2.25 (2)(2) Not in equilibriumShifts left!

Do an ICE chart CO (g) + H 2 O (g) ⇄ CO 2(g) + H 2(g) I 2.00 M2.00 M3.00 M3.00 M C+x+x-x-x E x x x x Keq=(3 - x) 2 =1.0 (2 + x) 2 Square root 3 - x = x 3 - x = 2 + x 1=2x x = 0.50 M [CO 2 ]=[H 2 ]= = 2.50 M [CO]=[H 2 O]= = 2.50 M

Size of the Keq

Big Keq Keq= Keq=10 products reactants

Little Keq- Keq= Keq=0.1 Note that the keq cannot be a negative number! products reactants

Keqabout 1 Keq= Keq=1 products reactants

Which reaction favours the products the most? Keq = 2.6 x 10 6 Keq = 3.5 x 10 2 Which reaction favours the reactants the most? Keq = 2.6 x Keq = 3.5 x 10 -7

Which reaction favours the products the most? Keq = 2.6 x 10 6 Keq = 3.5 x 10 2 Which reaction favours the reactants the most? Keq = 2.6 x Keq = 3.5 x 10 -7