Chemistry · Ch 12 — Chemical Equilibrium
Application of Equilibrium Constant
Application of Equilibrium Constant
The equilibrium constant has three main practical applications. (1) Predicting the direction of reaction: for , comparing the reaction quotient (Section 12.5) with tells us which way an unbalanced mixture will shift -- if , the reaction proceeds forward (left to right), generating more product; if , the system is already at equilibrium and no net reaction occurs; if , the reaction proceeds in reverse (right to left), generating more reactant. This comparison predicts only the DIRECTION of the shift, never how long it takes to get there. (2) Judging the extent of reaction: comparing (at 500 K, ) with its reverse, (at 500 K, ), shows that a very large () means a high proportion of products at equilibrium, with the forward reaction favoured and nearly going to completion, while a very small () means only a small fraction of reactants convert, with the reverse reaction favoured; values of in between mean appreciable amounts of both reactants and products are present at equilibrium. (3) Calculating equilibrium concentrations: given the initial amounts and , an initial/change/equilibrium composition table can be solved for the actual equilibrium concentrations, worked through in Problem 12.5 (H2/I2/HI) and the ester-formation example (Section 12.5's esterification system). (4) Linking equilibrium to kinetics: since (Section 12.4.3), a large implies (forward rate constant much larger, reaction nearly goes to completion), near one implies (comparable forward/reverse rate constants, appreciable amo …
Worked out. A diagram (printed in the book re-using the figure number '12.4', separate from the two rate/concentration-vs-time graphs earlier in the chapter also labelled Fig 12.4(a)/(b) -- a book numbering duplication disclosed for transparency) showing a reaction system with an arrow pointing toward 'Reactants -> Products' when Qc is less than Kc (movement toward equilibrium is in the forward direction), a balanced double-headed state when Qc equals Kc (reactants and products already at equilibrium, no net reaction), and an arrow pointing 'Products -> Reactants' when Qc is greater than Kc (movement toward equilibrium is in the reverse direction). The diagram visually summarises the rule: compare the current reaction quotient Qc to the true equilibrium constant Kc to predict …
What this figure shows. A horizontal scale showing the value of Kc plotted from 'very small' (far left, Kc around 10^-3 or below) through 1 (centre) to 'very large' (far right, Kc around 10^3 or above). At the very-small end the label reads 'reaction proceeds hardly at all' (almost entirely reactants at equilibrium); at the very-large end the label reads 'reaction proceeds nearly to completion' (almost entirely products at equilibrium); the central region is labelled 'appreciable concentrations of both reactants and products are present at equilibrium'. The diagram is the visual summary used to judge, from a Kc value alone, …
Worked out. Worked example: equal concentrations of H2 and I2 are mixed and allowed to reach equilibrium at 700 K, H2(g) + I2(g) is in equilibrium with 2HI(g), Kc = 54, and [HI] at equilibrium = 0.85 mol dm-3. Since the initial concentrations of H2 and I2 are equal and they react in a 1:1 ratio, their equilibrium concentrations remain equal to each other, call it x. Substituting into Kc = [HI]^2 / ([H2][I2]) gives 54 = (0.85)^2 / x^2, so x^2 = 0.7225/54 = 0.01338, and x = [H2] = [I2] = 0.12 mol …
Worked out. Worked example: starting with 2.0 mol ethanoic acid and 2.0 mol ethanol in V litres, and letting x mol of ethyl ethanoate form at equilibrium (Kc = 4.0 at the given temperature), the initial/equilibrium mole table reads -- initial: 2.0, 2.0, 0, 0; at equilibrium: (2.0-x), (2.0-x), x, x; equilibrium concentrations (dividing by V): (2.0-x)/V, (2.0-x)/V, x/V, x/V. Substituting into Kc = [ester][water]/([acid][alcohol]) gives 4.0 = (x/V)(x/V) / [(2.0-x)/V x (2.0-x)/V] = x^2/(2.0-x)^2, so 2 = x/(2.0-x), giving 4-2x = x, x = 4/3 = 1.33 mol, and (2.0-x) = 0.67 mol. Equilibrium composition: 0.67 mol each of ethanoic acid and ethanol, and 1.33 mol of ethyl ethanoate, in V litres. This ICE-style mole table was cross-checked arithmetically end to end and is internally consistent …
Worked out. Worked example for H2(g) + I2(g) is in equilibrium with 2HI(g), Kc = kf/kr = 54.0 at 700 K. Part (a): since Kc = kf/kr = 54.0, kf (rate constant for HI formation) is larger than kr (rate constant for HI decomposition) by a factor of 54.0. Part (b): given kr = 1.16 x 10^-3 at 700 K, kf = Kc x kr = 54.0 x (1.16 x 10^-3) = 62.64 x 10^-3, i.e. about 0.0626. This links the equilibrium constant directly to the ratio of the forward and reverse rate constants from chemic …