Important features of Equilibrium

APPLICATIONS OF  EQUILIBRIUM CONSTANTS

Before considering the applications of equilibrium constants, let us summarise the important features of equilibrium constants as follows:

1. Expression for equilibrium constant is applicable only when concentrations of the reactants and products have attained

constant value at equilibrium state.

2. The value of equilibrium constant is independent of initial concentrations of the reactants and products.

3. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

4. The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.

5. The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction, whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

Let us consider applications of equilibrium constant to:

(a) predict the extent of a reaction on the basis of its magnitude,

(b) predict the direction of the reaction, and

(c) calculate equilibrium concentrations.

Predicting the Extent of a Reaction

Predicting the Extent of a Reaction

The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction.But it is important to note that an equilibrium constant does not give any information about the rate at which the equilibrium is reached.The magnitude of K_c or K_p is directly proportional to the concentrations of products (as these appear in the numerator of equilibrium constant expression) and inversely proportional to the concentrations of the reactants (these appear in the denominator).This implies that a high value of K is suggestive of a high concentration of products and vice-versa.

We can make the following generalisations concerning the composition of

equilibrium mixtures:

1. If Kc > 10³, products predominate over reactants, i.e., if K_c is very large, the reaction proceeds nearly to completion.Consider the following examples:

(a) The reaction of H₂ with O₂ at 500 K has a very large equilibrium constant ,K_c =2.4 × 10⁴⁷.

(b) H₂ (g) + Cl₂ (g) 2HCl(g) at 300K has Kc = 4.0 × 10³¹.

(c) H₂(g) + Br₂(g)2HBr (g) at 300 K,K_c = 5.4 × 10¹⁸.

2. If K_c < 10^-3 , reactants predominate over products, i.e., if K_c is very small, the reaction proceeds rarely. Consider the following examples:

(a)The decomposition of H₂O into H₂ and O₂ at 500 K has a very small equilibrium constant, K_c = 4.1 × 10^–48.

(b)N₂(g) + O₂(g)2NO(g),at 298 K has K_c = 4.8 ×10^–31.

3. If K_c is in the range of 10 ^–3 to 10³, appreciable concentrations of both reactants and products are present.Consider the following examples:

(a) For reaction of H₂ with I₂ to give HI,K_c = 57.0 at 700K.

(b) Also, gas phase decomposition of N₂O₄ to NO₂ is another reaction with a value of K_c = 4.64 × 10^–3 at 25°C which is neither too small nor too large. Hence,equilibrium mixtures contain appreciable concentrations of both N₂O₄ and NO₂.

These generarlisations are illustrated in Fig. 7.6

Fig.7.6 Dependence of extent of reaction on K_c

Predicting the Direction of the Reaction

Predicting the Direction of the Reaction

The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage.For this purpose, we calculate the reaction quotient Q. The reaction  quotient,Q (Q_c with  molar concentrations and Q_P with partial pressures)is defined in the same way as the equilibrium constant Kc except that the concentrations in Q_c are not necessarily equilibrium values.For a general reaction:

a A + b Bc C + d D           (7.19)

Q_c = [C]^c[D]^d / [A]^a[B]^b           (7.20) 

^Then,

If Q_c > K_c, the reaction will proceed in the direction of reactants (reverse reaction).

If Q_c < K_c, the reaction will proceed in the direction of the products (forward reaction).

If Q_c = K_c, the reaction mixture is already at equilibrium.

Consider the gaseous reaction of H₂ with I₂,

H₂(g)+I₂(g)2HI(g);K_c = 57.0 at 700 K.

Suppose we have molar concentrations [H₂]_t=0.10M, [I₂]_t = 0.20 M and [HI]_t = 0.40 M.(the subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium).

Thus, the reaction quotient,Q_c at this stage of the reaction is given by,

Q = [HI]\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0baaSqaai % aadshaaeaacaaIYaaaaaaa!37D8! _t^2\) / [H₂]_t[I₂]_t= (0.40)²/ (0.10)×(0.20)=8.0

Now, in this case, Q_c (8.0) does not equal K_c (57.0), so the mixture of H₂(g), I₂(g) and HI(g) is not at equilibrium; that is, more H₂(g) and I₂(g) will react to form more HI(g) and their concentrations will decrease till Q_c = K_c.

The reaction quotient, Q_c is useful in predicting the direction of reaction by comparing the values of Q_c and K_c.

Thus,  we can  make  the  following

generalisations concerning the direction of the reaction (Fig. 7.7) :

Fig. 7.7 Predicting the direction of the reaction

1. If Q_c< K_c , net reaction goes from left to right

2. If Q_c > K_c , net reaction goes from right to left.

3. If Q_c > K_c , no net reaction occurs.

PROBLEM 7

The value of Kc for the reaction 2A BC is 2 × 10^–3. At a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 × 10^–4 M. In which

direction the reaction will proceed?

SOLUTION

For the reaction the reaction quotient Q_c is given by,

Q_c = [B][C]/ [A]²

as [A] = [B] = [C] = 3 × 10^–4M

Q_c = (3 ×10 )(3 × 10 ) / (3 ×10 ) = 1

as Q_c > K_c so the reaction will proceed in the reverse direction.

Calculating Equilibrium Concentrations

Calculating Equilibrium Concentrations

In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:

Step 1. Write the balanced equation for the reaction.

Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction:

a. the initial concentration,

b. the change in concentration on going to equilibrium, and

c. the equilibrium concentration.

In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x.

Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.

Step 4. Calculate the equilibrium concentrations from the calculated value of x.

Step 5. Check your results by substituting them into the equilibrium equation.

PROBLEM 8

13.8g of N₂O₄ was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium N₂O₄ (g)2NO₂ (g).The total pressure at equilbrium was found to be 9.15 bar. Calculate K_c, K_p and partial pressure at equilibrium.

SOLUTION

We know pV = nRT

Total volume (V ) = 1 L Molecular mass of N₂O₄ = 92 g

Number of moles = 13.8g/92 g = 0.15

of the gas (n)

Gas constant (R) = 0.083 bar L mol^–1K^–1 Temperature (T ) = 400 K

pV = nRT

p × 1L = 0.15 mol × 0.083 bar L mol^–1K^–1× 400 K

p = 4.98 bar

N₂O₄ 2NO₂

Initial pressure: 4.98 bar             

At equilibrium: (4.98 – x) bar   2x bar Hence,

p_total at equilibrium =\( {p_{{N_2}{O_4}}} + {p_{N{O_2}}}\)

9.15 = (4.98 – x) + 2x

9.15 = 4.98 + x

x = 9.15 – 4.98 = 4.17 bar

Partial pressures at equilibrium are,

\({p_{{N_2}{O_4}}}\)= 4.98 – 4.17 = 0.81bar

\({p_{N{O_2}}}\) = 2x = 2 × 4.17 = 8.34 bar

\( {K_p} = {\left( {{p_{N{O_2}}}} \right)^2}/{p_{{N_2}{O_4}}}\)

= (8.34)²/0.81 = 85.87

\({K_p} = {K_c}{\left( {RT} \right)^{\Delta n}}\)

85.87 = K_c (0.083 × 400)¹

K_c = 2.586 = 2.6

PROBLEM 9

3.00 mol of PCl₅ kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the mixture at equilibrium. K_c= 1.80.

SOLUTION

Initial PCl₅PCl₃+Cl₂

concentration: 3.0              0            0

Let x mol per litre of PCl₅ be dissociated,

At equilibrium:(3x-x)         x              x

K_c = [PCl₃][Cl₂]/[PCl₅]

1.8 = x²/ (3 – x)

x² + 1.8x – 5.4 = 0

x = [–1.8 ± (1.8)² – 4(–5.4)]/2

x = [–1.8 ± 3.24 + 21.6]/2 x = [–1.8 ± 4.98]/2

x = [–1.8 + 4.98]/2 = 1.59

[PCl₅] = 3.0 – x = 3 –1.59 = 1.41 M [PCl₃] = [Cl₂] = x = 1.59 M