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Chemistry 1011. TOPIC Gaseous Chemical Equilibrium TEXT REFERENCE Masterton and Hurley Chapter 12. 12.2 The Equilibrium Constant. YOU ARE EXPECTED TO BE ABLE TO: Write an expression for the equilibrium constant, K, for a gaseous reaction - PowerPoint PPT Presentation
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Chemistry 1011 Slot 5 1
Chemistry 1011
TOPICGaseous Chemical Equilibrium
TEXT REFERENCEMasterton and Hurley Chapter 12
Chemistry 1011 Slot 5 2
12.2 The Equilibrium Constant
YOU ARE EXPECTED TO BE ABLE TO:
• Write an expression for the equilibrium constant, K, for a gaseous reaction
• Recognize that the expression for K depends on the form of the balanced chemical equation for the reaction.
• Write an expression for the equilibrium constant, K, for a gaseous reaction that includes a substance in the solid or liquid phase.
Chemistry 1011 Slot 5 3
The Equilibrium Constant
• For a gaseous reaction, the equilibrium constant can be written in terms of the partial pressures (concentrations) of reactants and products
• For aA(g) + bB (g) cC (g) + dD (g)
• The equilibrium constant, Kp , is
K = (PC)c x (PD)d
(PA)a x (PB)b
Chemistry 1011 Slot 5 4
The Equilibrium Constant
• Equilibrium partial pressures of the products are in the numerator (top)
• Equilibrium partial pressures of the reactants are in the denominator (bottom)
• Each partial pressure is raises to a power equal to its coefficient in the balanced equation
Chemistry 1011 Slot 5 5
Equilibrium Constant Example
• Ammonia is made industrially by the Haber Process:
N2(g) + 3H2(g) 2NH3(g)
• The equilibrium constant, K, is
Kp = (PNH3)2
PN2 x (PH2
)3
Chemistry 1011 Slot 5 6
Equilibrium Constant Example
• Sulfuric acid is a very important industrial chemical. It is manufactured from sulfur dioxide and oxygen
2SO2(g) + O2(g) 2SO3(g)
• The equilibrium constant, K, is
Kp = (PSO3
)2
(PSO2 )2 x PO2
Chemistry 1011 Slot 5 7
Dependence of K on Equation Stoichiometry
• The expression for K, and its value will depend on how the equation is written
• For N2(g) + 3H2(g) 2NH3(g)
Kp = (PNH3)2
PN2 x (PH2
)3
• For 1/2N2(g) + 3/2H2(g) NH3(g)
Kp’ = (PNH3)
(PN2 )1/2 x (PH2
)3/2
Chemistry 1011 Slot 5 8
Dependence of K on Equation Stoichiometry
• The Coefficient Rule:– If coefficients in a balanced equation are
multiplied by a factor, n, then– The equilibrium constant is raised to the nth
power
K’ = Kn
• The Reciprocal Rule:– If the equation is written in reverse, then
K’’ = 1/K
Chemistry 1011 Slot 5 9
Adding Chemical Equations
• The Rule of Multiple Equilibria
• If a reaction can be expressed as the sum of two or more reactions, K for the overall reaction is equal to the PRODUCT of the equilibrium constants for the individual reactions– If Reaction 3 = Reaction 1 + Reaction 2
– Then K(Reaction 3) = K(reaction 1) x K(Reaction 2)
Chemistry 1011 Slot 5 10
Multiple Equilibria ExampleReaction 1: SO2(g) + 1/2O2(g) SO3(g)
Kp = 2.2 = (PSO3)
(PSO2 ) x (PO2
)1/2
Reaction 2: NO2(g) NO(g) + 1/2O2(g)
Kp = 4.0 = (PNO) x (PO2 )1/2
(PNO2 )
Adding the equations:
SO2(g) + NO2(g) SO3(g) + NO(g)
Chemistry 1011 Slot 5 11
Multiple Equilibria Example• The equilibrium constant expression for the
total reaction is Kp = (PSO3
) x (PNO)
(PSO2 ) x (PNO2
)
This is obtained by multiplying together the equilibrium constant expressions for the two individual reactions
(PSO3) x (PNO) x (PO2
)1/2
(PSO2 ) x (PO2
)1/2 (PNO2 )
Kp = 2.2 x 4.0 = 8.8
Chemistry 1011 Slot 5 12
Heterogeneous Equilibria
• Up to now, all of the reactions considered have been homogeneous gas reactions
• In some cases, one or more of the substances involved may be a liquid or solid
• This would be a heterogeneous system
• Example:
I2(s) I2(g)
Chemistry 1011 Slot 5 13
Heterogeneous Equilibria – the Sublimation of Iodine
I2(s) I2(g)
• The rate of the forward process depends only on temperature. Sublimation will occur at constant rate as long as some solid iodine remains
• The rate of the reverse reaction will depend on the concentration (partial pressure) of iodine gas
• At equilibrium (in a closed system), this rate will become constant
Kp = PI2(g)
Chemistry 1011 Slot 5 14
Heterogeneous Equilibria
• For heterogeneous equilibria, it is found that:– The position of equilibrium is independent of
the amount of solid or liquid component, as long as some still remains
– Concentrations of solids and liquids are constant
– Terms for solid or liquid components do not appear in the expression for K
Chemistry 1011 Slot 5 15
Heterogeneous Equilibria – the Decomposition of Calcium Carbonate
CaCO3(s) CaO(s) + CO2(g)
• Written in terms of concentrations, the equilibrium constant expression is:
K = [CaO(s)][CO2(g)]
[CaCO3(s)]• But the concentrations of the solids are
constant, so
K = [CO2(g)], or Kp = PCO2
Chemistry 1011 Slot 5 16
Heterogeneous Equilibria with a Liquid Component
• Example:
CO2(g) + H2(g) CO(g) + H2O(l)
• In this case, the amount of water vapour in the system is constant, since it will be in equilibrium with the liquid water
• So, Kp = (PCO)
(PCO3) (PH3
)