Reactivity 2.3.5—The reaction quotient, Q, is calculated using the equilibrium expression with nonequilibrium concentrations of reactants and products. (AHL)
Reactivity 2.3.6—The equilibrium law is the basis for quantifying the composition of an equilibrium mixture. (AHL)
Reactivity 2.3.7—The equilibrium constant and Gibbs energy change, ΔG, can both be used to
measure the position of an equilibrium reaction. (AHL)
Reaction Quotient Q and Equilibrium Constant Kc
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Consider the following reversible reaction: mA + nB ⇌ pC + qD
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The Reaction Quotient measures the relative amount of products and reactants present during a reaction at a particular point in time:
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Q = Kc → the reaction has reached dynamic equilibrium
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Q>Kc → the concentration of products is greater than at equilibrium and the reverse reaction is favored to reach Kc
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Q<Kc → the concentration of reactants is greater than at equilibrium and the forward reaction is favored to reach Kc
Calculating Kc
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Using ICE table, find equilibrium concentration values of reactants and products
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Kc doesn’t have units
Scenario 1 - Calculating the concentrations at equilibrium (standard) |
A student placed 0.40 mol of PCl3(g) and 0.20 mol of Cl2(g) into a 1.0 dm3 flask at 250 °C. The reaction, which produced PCl5, was allowed to come to equilibrium, at which time it was found that the flask contained 0.05 mol of Cl2. What is the value of Kc for this reaction? |
*values in black are given in the question Step 1
• Identify the change for Cl2 (-0.15)Step 2
• Calculate the change for PCl3 as identical to Cl2, i.e. -0.15 (in accordance to 1:1 coefficient ratio)
• Calculate the change for PCl5 as opposite of Cl2, i.e. +0.15Step 3
• Calculate the equilibrium (mol) by applying the changes made to the initial (mol)Step 4
• Find actual [equilibrium] valuesStep 5
• Kc = [PCl5][PCl3][Cl2]= [0.15][0.25][0.05] = 12 |
Scenario 2 - Calculating the concentrations at equilibrium (extended) |
A closed vessel with volume of 500cm3 initially contained 0.4 moles of both hydrogen and nitrogen, the reaction was allowed to reach equilibrium where 0.1 moles of hydrogen remained. Calculate the Kc and its units. Assume that the temperature and volume of the container remained unchanged. |
*values in black are given in the question Step 1
• Identify the change for H2 (-0.3)Step 2
• Calculate the change for N2 as 3 times less than H2, i.e. -0.1 (in accordance to 1:3 coefficient ratio)
• Calculate the change for NH3 as opposite and ⅔ of H2 (or twice bigger than N2), i.e. +0.2Step 3
• Calculate the equilibrium (mol) by applying the changes made to the initial (mol)Step 4
• Find [equilibrium] by dividing by volume (0.5 dm3)Step 5
• Kc = [NH3]2[N2][H2]3= [0.4]2[0.6][0.2]3 = 33.3 |
N.B.// [equilibrium] value derivation may be skipped b/c Kc value calculated from equilibrium (mol) or equilibrium [equilibrium] are the same
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When the volume of the container is 1.00 mol dm-3
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When the reaction is equimolar (equal no. of moles on each side) e.g. I2 + H2 ⇌ 2HI
Scenario 3 - Calculating the concentrations at equilibrium when given the Kc
The Kc for this reaction at 700K is 0.0184. What are the concentrations at equilibrium when starting with 0.100 mol for each reactant? The volume of the vessel is 500cm3.
Step 1 | • Identify the change for product with assumption that 2x moles of product remained |
Step 2 | • Calculate the change for H2 and I2 as opposite and twice less, i.e. -x (in accordance to 1:2 coefficient ratio) |
Step 3 | • Calculate the equilibrium (mol) by applying the changes made to the initial (mol) |
Step 4 | • Since reaction is equimolar, directly use equilibrium (mol) values as the concentration values
Kc = [HI]2[H2][I2] = 0.0184[2x]2[0.100-x]2 = 0.01842x0.100-x = 0.0184=0.1362x = 0.136 (0.100-x) 2x=0.0136-0.136x2.136x=0.0136x=6.3710-3 |
Step 5 | • Calculate the concentrations by (moles final) / volume (0.5 dm3)
• Moles final of H2 & I2 Moles final of HI0.100 -x=0.100-(6.3710-3) = 0.094 mol2x= 2 (6.3710-3) = 0.0127 mol0.0940.5=0.188 mol dm-30.0127 0.5 = 0.0254 mol dm-3 |
Kc for the inverse/multiple of a reaction
cC + dD ⇌ aA + bB
Kc=[A]a[B]b[C]c[D]d
Kc′= 1/Kc or Kc′ = Kc–1
2aA + 2bB ⇌ 2cC + 2dD
Kc=[C]2c[D]2d[A]2a[B]2b
Kc′= Kc2 (Kcx)
Gibbs Free Energy and Equilibrium
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∆G=∆Gϴ +RT lnQ
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∆Gϴ= -RT lnK
ΔG of forward reaction | Position of Equilibrium |
Negative | Forward reaction is favoured |
Positive | Reverse reaction is favoured |
Zero | Neither side is favoured/favoured equally |
e.g. The reaction X ⇌ Y has a ΔG value of -92.3kJmol-1. Calculate Kc and deduce which side is favoured at 300K.
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Value of R is given in data booklet R= - 8.31JK-1mol-1
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92300=-8.31300 lnK
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lnK= -92300-8.31300=37.00
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K=e37.00=1.18106
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Since K > 1, equilibrium lies to the right