Chapter 14- Chemical Equilibrium
Homogenous Equilibrium
Applies to reactions in which all of the reacting species are in the same phase
Heterogenous Equilibrium
Applies to reactions in which reactants and products are in different phases
Reaction Quotient (Qc)
Calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression
What does it mean if the equilibrium constant is less than 1?
Equilibrium will lie to the left, favoring the reactants
What does it mean if the equilibrium constant is greater than 1?
Equilibrium will lie to the right, favoring the products
A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00-L stainless-steel flask at 430°C The equilibrium constant Kc for the reaction H2(g) + I2(g) --->2HI(g) is 54.3 at this temperature Calculate the concentrations of H2, I2, and HI at equilibrium
H2 I2 HI I 0.500 0.500 0.000 C -x -x +2x E 0.500-x 0.500-x 2x ------------------------------- 54.3= (2x)^2/(0.500-x)^2 7.3688= 2x/(0.500-x) 3.684 - 7.3688x = 2x 3.684 = 9.3688x x= 0.39322 ------------------------------- [H2], [12]= 0.500-0.39322= 0.106M [HI]= 0.78644
Equilibrium Constants
Kc=[product]^x[product]^y/[reactant]^a[reactant]^b
What is the formula for Kp?
Kp= Kc(RT)^delta n
What are the units for Kp and Kc?
NONE
When can we determine Kp?
ONLY IF THERE ARE GASES PRESENT
What are not included in the equilibrium constant expression?
Pure solids and liquids
For the same reaction and temperature as in the previous example, H2(g) + I2(g)---> 2HI(g) suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively Calculate the concentrations of these species at equilibrium
Qc= (0.0224)^2/ (0.00623)(0.00414) Qc= 19.45 --------------------------------- H2 I2 HI I 0.00623 0.00414 0.0224 C -x -x +2x E (0.00623-x) (0.00414-x) (0.0224+2x) ------------------------------- 1)54.3= (0.0224+2x)^2/(0.00623-x)(0.00414-x) 2) 54.3(2.58x10^-5 - 0.0104x +x^2)= 5.02x10^-4 +0.0896x +4x^2 3) 50.3x^2 - 0.654x +8.98x10^-4 4) x= 0.654 +/- sqrt (0.654^2)-(4)(50.3)(8.98x10^-4) all over 100.6 5) x= 0.654 +/- 0.4986 all over 100.6 6) x= 0.011457 or x= 0.00124 FINAL X= 0.00124 ------------------------------- [H2]= 0.00623- 0.00124= 0.00499 [I2]= 0.00414- 0.00124= 0.0029 [HI]= 0.0224+0.00248= 0.02488
Equilibrium
The state in which there are no observable changes as time goes by
If Qc is equal to Kc then...
The system is at equilibrium
If Qc is less than Kc then...
The system proceeds from left to right to reach equilibrium
If Qc is greater than Kc then...
The system proceeds from right to left to reach equilibrium
At the start of a reaction, there are 0.249 mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50-L reaction vessel at 375°C. If the equilibrium constant (Kc) for the reaction N2(g) + 3H2(g)---> 2NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.
[N2]= 0.249/3.5= 0.0711M [H2]= 0.0321/3.5=0.0092M [NH3]=0.000642/3.5=0.000183M ------------------------------- Qc= (0.000183)^2/ (0.0711)(0.0092)^3 Qc= 0.6088 ------------------------------ Qc < Kc, so the reaction is not in equilibrium, the reaction will proceed from left to right
Consider the following heterogeneous equilibrium: CaCO3(s) --->CaO(s) + CO2(g) At 800°C, the pressure of CO2 is 0.236 atm. Calculate (a) KP and (b) Kc for the reaction at this temperature
a) Kp= P(CO2) FINAL ANSWER= 0.236 atm b) Kp= Kc(0.082 x 1073)^1 0.236=Kc(87.986) Kc= 0.00268 FINAL ANSWER= 0.00268
When is chemical equilibrium achieved?
-The rates of the forward and reverse reactions are equal -Concentrations of the reactants and products remain constant
Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) -->CH3OH(g) The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of Kp at this temperature?
1) Kp= 10.5(0.082)(493)^-2 FINAL ANSWER = 0.0064
The equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g) PCl3(g) + Cl2(g) is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250°C?
1)Kp= P(PCL3) 2)(P(Cl2))/(P(PCl5)) 3)1.05= (0.463)(P(Cl2))/0.875 4)0.91875=0.463(P(Cl2)) 5)P(Cl2)= 1.984 atm FINAL ANSWER: 1.984
