Equilibrium

NCERT Solution

Part 1

Question 1: A liquid is in equilibrium with its vapor in a sealed container at a fixed temperature. The volume of container is suddenly increased.

(a) What is the initial effect of the change on vapor pressure?

Answer: No change

(b) How do rates of evaporation and condensation change initially?

Answer: Rate of evaporation increases while rate of condensation decreases.

(c) What happens when equilibrium is restored finally and what will be the final vapor pressure?

Answer: Vapor pressure remains the same as it does not depend on volume but on temperature.


Question 2: What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2] = 0.60 M, [O2 = 0.82 M and [SO3] = 1.90 M?

2SO2 (g) + O2 ⇌ 2SO3 (g)

Answer: `K_c=([SO_3]^2)/([SO_2]^2[O_2])`

`=(1.90^2)/(0.60^2xx0.82)`

`=(3.61)/(0.36xx0.82)=12.20`

Question 3: At a certain temperature and total pressure of 105 Pa, iodine vapor contains 40% by volume of I atoms

I2 (g) ⇌ 2I (g)

Calculate Kp for the equilibrium.

Answer: Partial pressure of Iodine atoms = 105 × 40% = 4 × 104 Pa

Partial Pressure of Iodine molecules = 105 × 60% = 6 × 104 Pa

`K_p=((4xx10^4Pa)^2)/(6xx10^4Pa)`

= 2.6 × 104 Pa

Question 4: Write the expression for the equilibrium constant Kc for each of the following reactions:

(a) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g)

Answer: `([NO]^2[Cl_2])/([NOCl]^2)`

(b) 2Cu(NO3)2 (s) ⇌ 2CuO (s) + 4NO2 (g) + O2 (g)

Answer: `([CuO]^2[NO_2]^4[O_2])/([Cu(NO)_3]^2)`

(c) CH3COOC2H5 (aq) + H2 (l) ⇌ CH3COOH (aq) + C2H5OH (aq)

Answer: `([CH_3COOH][C_2H_5OH])/([CH_3COOC_2H_5][H_2])`

(d) Fe3+ (aq) + 2OH- (aq) ⇌ Fe(OH)3 (s)

Answer: `([Fe(OH)_3])/([Fe^(3+)][OH^-]^2)`

(e) I2 (s) + 5F2 ⇌ 2IF5

Answer: `([IF_5]^2)/([I_2][F_2]^5)`

Question 5: Find out the value of Kc for each of the following equilibria from the value of Kp:

(a) 2NOCl (g) ⇌ 2NO (g) + Cl2 (g): Kp = 1.8 × 10-2 at 500 K

Answer: We know, Kp = Kc (RT)Δng

Kp = 1.8 × 10-2 atm

Δng = 3 – 2 = 1

R = 0.0821 L atm K-1 mol-1

T= 500 K

So, `K_c=(K_p)/((RT)^(Δng))`

`=(1.8xx10^(-2))/(0.0821xx500)`=

= 4.4 × 10-4 mol L-1

(b) CaCO3 (s) ⇌ CaO (s) + CO2: Kp = 167 at 1073 K

Answer: `K_c=(K_p)/((RT)^(^Delta;ng))`

`=(167)/(0.0821xx1073)=1.9` mol L-1

Question 6: For the following equilibrium, Kc = 6.3 × 1014 at 1000 K

NO (g) + O3 ⇌ NO2 (g) + O2 (g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc for the reverse reaction?

Answer: Kc for reverse reaction will be reciprocal of Kc for forward reaction.

`K_c=1/(6.3xx10^(14))`

`=(10^(-14))/(6.3)=1.6xx10^(-13)`


Question 7: Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Answer: Volumes of pure liquids and solids generally remain constant. So, they can be ignored while writing the equilibrium constant expression.

Question 8: Reaction between N2 and O2, takes place as follows:

2N2 (g) + O2 (g) ⇌ 2N2O (g)

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 × 10-37, determine the composition of equilibrium mixture.

Answer: We know that if Kc < 10-3 then reaction rarely proceeds. Here, Kc is much less than that. So, the reaction will rarely proceed.

Let us assume that x moles of N2 take part in reaction. According to equation, `x/2` moles of O2will react with this to form x mole of N2O.

2N2 (g)O2 (g)2N2O (g)
Initial mole L-1`(0.482)/(10)``(0.933)/(10)`Zero
Mole L-1 at equilibrium`(0.482-x)/(10)``(0.933-x/2)/(10)``x/(10)`

Now, applying the Law of Equilibrium we get following equation:

`K_c=([N_2O]^2)/([N_2]^2[O_2])`

Or, `2.0xx10^(-7)=((x/(10))^2)/(((0.482)/(10))^2xx((0.933)/(10)))`

`=(0.01x^2)/(2.1676xx10^(-4))`

`x^2=43.352xx10^(-40)`

Or, `x=6.6xx10^(-20)`

Equilibrium mixture:

Molar concentration of N2 = 0.0482 mol L-1

Molar concentration of O2 = 0.0933 mol L-1

Molar concentration of N2O = `0.1xx\x=0.1xx6.6xx10^(-20)` mol L-1

= 6.6 × 10-21 mol L-1

Question 9: Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO (g) + Br2 (g) ⇌ 2NOBr (g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

Answer: According to balanced equation, 2 mole of NO reacts with 1 mole of Br2 to give 2 mole of NOBr.

So, if 0.0518 M of NOBr is participating in reaction then same amount of NO is participating and half of it (0.0259 M) of Br2 is participating in reaction.

Original amount of NO = 0.087

Amount at equilibrium = 0.087 – 0.0518 = 0.0352 M

Amount of Br2 at equilibrium = 0.0437 – 0.0259 = 0.0178 N

Amount of NOBr at equilibrium = 0.0518 (given)

Question 10: At 450K, Kp = 2.0 × 1010 bar for the given reaction at equilibrium. What is Kc at this temperature?

2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

Answer: We know, Kp = Kc(RT)Δng

R = 0.083 L bar K-1 mol-1, T = 450 K and Δng = 2 – 3 = - 1

So, Kc `=(2.0xx10^(10))/((0.083xx450)^(-1))`

`=2.0xx10^(10)xx0.083xx450=7.47xx10^(11)` M-1

Question 11: A sample of HI (g) is placed in a flask at pressure of 0.2 atm. At equilibrium the partial pressure of HI (g) is 0.04 atm. What is Kp for the given equilibrium?

2HI (g) ⇌ H2 (g) + I2 (g)

Answer: Initial pressure of HI = 0.2 atm

Pressure of HI at equilibrium = 0.04

So, reduction in pressure = 0.2 – 0.04 = 0.16

This will be the total pressure of mixture of H2 and I2

So, pressure of H2 = `(0.16)/2=0.08`

Pressure of I2 will be same, i.e. 0.08

Now Kp can be calculated as follows:

`K_p=(0.08xx0.08)/(0.04xx0.04)=4.0` atm


Question 12: A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer: Concentration quotient of equation can be calculated as follows:

`Q_c=([NH_3]^2)/([N_2][H_2]^3)`

`=((8.13)/(20))/((1.57)/(20)((1.92)/(20))^3)`

`=(8.13xx20xx20xx20xx20)/(20xx1.57xx1.92xx1.92xx1.92)`

`=(8.13xx8000)/(11.11)=6xx10^3`

Here, Qc ≠ Kc

So, the reaction mixture is not at equilibrium.

Since Qc > Kc, the reaction will shift to right.

Question 13: The equilibrium constant expression for a gas reaction is

`K_c=([NH_3]^4[O_2]^5)/([NO]^4[H_2O]^6)`

Write the balanced chemical equation corresponding to this expression.

Answer: 4NO (g) + 6H2O (g) ⇌ 4NH3 (g) + 5O2 (g)

Question 14: One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation.

H2O (g) + CO (g) ⇌ H2(g) + CO2 (g)

Calculate the equilibrium constant for the reaction.

Answer: As per balanced equation, 1 M of H2 reacts with 1 M of CO to give 1 M of each product. (Per Liter Reaction Mixture)

So, 40% of 1 M = 0.4 M of H2 will react with 0.4 M of CO to give 0.4 M of each product.

This means, 0.6 M of each reactant is left at equilibrium.

Hence, equilibrium constant can be calculated as follows:

`K_c=([H_2][CO_2])/([H_2O][CO])`

`=(0.4xx0.4)/(0.6xx0.6)=0.44`

Question 15: At 700 K, equilibrium constant for the following reaction is 54.8.

H2 (g) + I2 (g) ⇌ 2HI (g)

If 0.5 mol L-1 of HI (g) is present at equilibrium at 700 K, what are the concentration of H2 and I2 (g) assuming that we initially started with HI (g) and allowed it to reach equilibrium at 700 K?

Answer: `K_c=([HI]^2)/([H_2][I_2])`

Or, `54.8=(0.5^2)/(x^2)` If x is the concentration of each reactant.

Or, `x^2=(0.25)/(54.8)=0.00456`

Or, `x=sqrt(0.00456)`



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