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Chemical Reactions 2: Equilibrium & Oxidation-Reduction. Equilibrium Constant. The equilibrium constant (K c ) is defined as the ratio between the concentrations of the products and the reactants, each raised to the power corresponding to its coefficient in the balanced equation. - PowerPoint PPT Presentation
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Chemical Reactions 2: Equilibrium & Oxidation-Reduction
Equilibrium Constant
The equilibrium constant (Kc) is defined as the ratio between the concentrations of the products and the reactants, each raised to the power corresponding to its coefficient in the balanced equation.
aA(g) + bB(aq) cC(g) + dD(aq)Products
Reactants
Equilibrium Constant
The equilibrium constant (Kc):
_ONLY changes value if we change Temperature
_Does NOT have units
_ONLY includes gaseous or aqueous species at equilibrium
_Solid and Liquids are not included in Kc’s expression as their concentrations do not change
_If Kc > 1 (Product favoured equilibrium)
_If Kc < 1 (Reactants favoured equilibrium)
Equilibrium Constant
Write the equilibrium constant (Kc) for the following reactions:
a) CO(g) + H2O(g) CO2(g) + H2(g)
b) 2NO2(g) 2NO(g) + O2(g)
c) 4HCl(g) + O2(l) 2H2O(g) + 2Cl2(g)
d) CaCO3(s) CaO(s) + CO2(g)
𝑲 𝒄=[CO 2 ]❑ ⌈H 2 ⌉❑
[ CO ]❑ [ H 2O ]❑
𝑲 𝒄=[ NO ]𝟐❑⌈O 2⌉❑
[ NO2 ]𝟐
𝑲 𝒄=[ H2O ]𝟐❑ [ Cl2 ]𝟐❑
[ HCl ]𝟒
𝑲 𝒄=[𝐂𝐎𝟐 ]
Equilibrium Constant
Calculate the equilibrium constant of the reaction below, if we know that at a given temperature the concentration of each substance is:
[NO] = 0.062 M, [H2] = 0.012 M, [N2] = 0.019 M and [H2O] = 0.138 M
2NO(g) + H2(g) N2(g) + 2H2O(g)
653.7
Equilibrium Constant
Calculate the concentration of C2H4 at the equilibrium described by the reaction below, if you know that the equilibrium constant is 300, and the concentration of the other substances are:
[C2H5OH] = 0.1500 M and [H2O] = 0.0225 M
C2H4(g) + H2O(g) C2H5OH(g)
= 0.0222M
Equilibrium Constant
We inject 1 mole of N2O4 into a 2-L container, and raise the temperature to 120°C. When equilibrium is reached, the gaseous mixture contains 0.4 moles of N2O4. Calculate the value of Kc. (N2O4(g) 2NO2(g))
Initially 1 mole 0 mole
Conversion 1-x 2x (1:2 ratio)
Equilibrium 0.4 mole 1.2 mole (x = 0.6 mole)
Concentration 0.4 mole/2L = 0.2M 1.2 mole/2L = 0.6M
Kc = [NO2]2 / [N2O4] Kc = (0.6M)2 / (0.2M) Kc = 1.8
Equilibrium Constant
276g of NO2 injected in a 4-L container, and temperature elevated to 120°C. When equilibrium is reached, the gaseous mixture contains 1.2 moles of NO2. Calculate the value of Kc. (2NO2(g) N2O4 (g))
Initially 6 mole (NO2= 46 g/mol) 0 mole
Conversion 6-x x/2 (2:1 ratio)
Equilibrium 1.2 mole 2.4 mole (x = 4.8 mole)
Concentration 1.2 mole/4L = 0.3M 2.4 mole/4L = 0.6M
Kc = [N2O4] / [NO2]2 Kc = (0.6M) / (0.3M)2 Kc = 6.67
Equilibrium Constant
A technician introduces 2.5 moles of PCl5 into a 2-L container, and temperature elevated to 250°C. At this point Kc is 0.042. Calculate the concentration of each substance at equilibrium. (PCl5(g) PCl3(g) + Cl2(g))Initially 1.25M (2.5 moles/2L) 0M 0M
Conversion -x + x + x
Equilibrium 1.25 - x x x
Kc = [PCl3][Cl2] / [PCl5] 0.042 = x2 / (1.25 – x) [PCl5] = 1.04Mx2 + 0.042x – 0.0525 = 0 [PCl3] = 0.209MUsing x = (- b ± √b2-4ac) / 2a [Cl2] = 0.209Mx = - 0.251 (impossible) or x = 0.209
Equilibrium Constant
The solubility product constant (Ksp) is defined as the product of the concentrations of the dissolved ions, each raised to the power corresponding to its coefficient in the balanced equation.
aA(s) cC+(aq) + dD-
(aq)
Equilibrium Constant
The solubility product constant (Ksp):
_ONLY changes value if we change Temperature
_Does NOT have units
_ONLY includes aqueous species (ions) at equilibrium
_Dissolution equation must be balanced (charges*)
_If Ksp > 1 (very soluble)
_If Ksp < 1 (rather insoluble)
Equilibrium Constant
Equilibrium Constant
Write the solubility product constant (Ksp) for the following reactions:
Equilibrium Constant
_The solubility of a compound is the maximum concentration of solute that the solvent can dissolve (saturated solution)
_Solubility ONLY changes with temperature
_The solubility of a given solid compound determines the value of the solubility product constant at a given temperature
Equilibrium Constant
Equilibrium Constant
Rank the following salts in increasing order of solubility:
ZnS, CuS, FeS, CdS, MgCO3, CaCO3
CuS, CdS, ZnS, FeS, CaCO3, MgCO3
Equilibrium Constant
Determine the value of Ksp of silver chloride if, at 25°C its solubility is 1.3 x 10-5 M.
AgCl(s) Ag+(aq) + Cl-
(aq)
Ksp = [Ag+][Cl-]Ksp = (1.3 x 10-5)(1.3 x 10-5)
Ksp = 1.69 x 10-10
Equilibrium Constant
Determine the value of Ksp of CaF2 if, at 25°C its solubility is 3.5 x 10-4 M.
Ksp = [Ca2+][F-]2
Ksp = (3.5 x 10-4)(2*3.5 x 10-4)2
Ksp = 1.71 x 10-10
CaF2(s) Ca2+(aq) + 2F-
(aq)
Equilibrium Constant
Determine the value of Ksp of SrF2 if, at 25°C its solubility is 0.0073g in 100 ml.
MM(SrF2) = (1*88) + (2 * 19) = 126 g/mol 126g _____1 mol 0.0073g
______X X = 5.79 x 10-5 mol C = n/V = 5.79 x 10-5 mol / 0.1 L = 5.79 x 10-4 mol/L
Ksp = [Sr2+][F-]2
Ksp = (5.79 x 10-4)(2*5.79 x 10-4)2
Ksp = 7.76 x 10-10
SrF2(s) Sr2+(aq) + 2F-
(aq)
Equilibrium Constant
Precipitation: Separation technique based on the solubility of a given compound. Uses “common ion effect”.
aA(s) cC+(aq) + dD-
(aq)
_Increasing the concentration of either C+ or D-, will shift the equilibrium toward the precipitation of A(s).
_Soluble compounds containing the common ion (C+ or D- ) will be used to achieve the separation of the solid.
_Precipitation is a technique used to determine ions present in solution
Equilibrium Constant
You have two colourless solutions of PbCl2 and AgCl. Which of the following solutions would you rather use to identify the PbCl2 using the common ion effect? Explain your reasons.
_NaNO3
_HCl
_PbI2
Common ion Pb2+ would increase its concentration in solution, displacing the equilibrium toward precipitation of PbCl2. HCl is not useful because it could also precipitate AgCl, and we are only interested in PbCl2.