# 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 K_{c} for the following equilibrium when the equilibrium concentration of each substance is: [SO_{2}] = 0.60 M, [O_{2} = 0.82 M and [SO_{3}] = 1.90 M?

2SO_{2} (g) + O_{2} ⇌ 2SO_{3} (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 10^{5} Pa, iodine vapor contains 40% by volume of I atoms

I_{2} (g) ⇌ 2I (g)

Calculate K_{p} for the equilibrium.

**Answer:** Partial pressure of Iodine atoms = 10^{5} × 40% = 4 × 10^{4} Pa

Partial Pressure of Iodine molecules = 10^{5} × 60% = 6 × 10^{4} Pa

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

= 2.6 × 10^{4} Pa

Question 4: Write the expression for the equilibrium constant K_{c} for each of the following reactions:

(a) 2NOCl (g) ⇌ 2NO (g) + Cl_{2} (g)

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

(b) 2Cu(NO_{3})_{2} (s) ⇌ 2CuO (s) + 4NO_{2} (g) + O_{2} (g)

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

(c) CH_{3}COOC_{2}H_{5} (aq) + H_{2} (l) ⇌ CH_{3}COOH (aq) + C_{2}H_{5}OH (aq)

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

(d) Fe^{3+} (aq) + 2OH^{-} (aq) ⇌ Fe(OH)_{3} (s)

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

(e) I_{2} (s) + 5F_{2} ⇌ 2IF_{5}

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

Question 5: Find out the value of K_{c} for each of the following equilibria from the value of K_{p}:

(a) 2NOCl (g) ⇌ 2NO (g) + Cl_{2} (g): K_{p} = 1.8 × 10^{-2} at 500 K

**Answer:** We know, K_{p} = K_{c} (RT)^{Δng}

K^{p} = 1.8 × 10^{-2} atm

Δ_{ng} = 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) CaCO_{3} (s) ⇌ CaO (s) + CO_{2}: K_{p} = 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, K_{c} = 6.3 × 10^{14} at 1000 K

NO (g) + O_{3} ⇌ NO_{2} (g) + O_{2} (g)

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_{c} for the reverse reaction?

**Answer:** K_{c} for reverse reaction will be reciprocal of K_{c} 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 N_{2} and O_{2}, takes place as follows:

2N_{2} (g) + O_{2} (g) ⇌ 2N_{2}O (g)

If a mixture of 0.482 mol N_{2} and 0.933 mol of O_{2} is placed in a 10 L reaction vessel and allowed to form N_{2}O at a temperature for which K_{c} = 2.0 × 10^{-37}, determine the composition of equilibrium mixture.

**Answer:** We know that if K_{c} < 10^{-3} then reaction rarely proceeds. Here, K_{c} is much less than that. So, the reaction will rarely proceed.

Let us assume that x moles of N_{2} take part in reaction. According to equation, `x/2` moles of O_{2}will react with this to form x mole of N_{2}O.

2N_{2} (g) | O_{2} (g) | 2N_{2}O (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 N_{2} = 0.0482 mol L^{-1}

Molar concentration of O_{2} = 0.0933 mol L^{-1}

Molar concentration of N_{2}O = `0.1xx\x=0.1xx6.6xx10^(-20)` mol L^{-1}

= 6.6 × 10^{-21} mol L^{-1}

Question 9: Nitric oxide reacts with Br_{2} and gives nitrosyl bromide as per reaction given below:

2NO (g) + Br_{2} (g) ⇌ 2NOBr (g)

When 0.087 mol of NO and 0.0437 mol of Br_{2} are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br_{2}.

**Answer:** According to balanced equation, 2 mole of NO reacts with 1 mole of Br_{2} 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 Br_{2} is participating in reaction.

Original amount of NO = 0.087

Amount at equilibrium = 0.087 – 0.0518 = 0.0352 M

Amount of Br_{2} at equilibrium = 0.0437 – 0.0259 = 0.0178 N

Amount of NOBr at equilibrium = 0.0518 (given)

Question 10: At 450K, K_{p} = 2.0 × 10^{10} bar for the given reaction at equilibrium. What is K_{c} at this temperature?

2SO_{2} (g) + O_{2} (g) ⇌ 2SO_{3} (g)

**Answer:** We know, K^{p} = K_{c}(RT)^{Δng}

R = 0.083 L bar K^{-1} mol^{-1}, T = 450 K and Δ_{ng} = 2 – 3 = - 1

So, K_{c} `=(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 K_{p} for the given equilibrium?

2HI (g) ⇌ H_{2} (g) + I_{2} (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 H_{2} and I_{2}

So, pressure of H_{2} = `(0.16)/2=0.08`

Pressure of I_{2} will be same, i.e. 0.08

Now K^{p} 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 N_{2}, 1.92 mol of H_{2} and 8.13 mol of NH_{3} is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_{c} for the reaction N_{2} (g) + 3H_{2} (g) ⇌ 2NH_{3} is 1.7 × 10^{2}. 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, Q_{c} ≠ K_{c}

So, the reaction mixture is not at equilibrium.

Since Q_{c} > K_{c}, 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) + 6H_{2}O (g) ⇌ 4NH_{3} (g) + 5O_{2} (g)

Question 14: One mole of H_{2}O 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.

H_{2}O (g) + CO (g) ⇌ H_{2}(g) + CO_{2} (g)

Calculate the equilibrium constant for the reaction.

**Answer:** As per balanced equation, 1 M of H_{2} 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 H_{2} 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.

H_{2} (g) + I_{2} (g) ⇌ 2HI (g)

If 0.5 mol L^{-1} of HI (g) is present at equilibrium at 700 K, what are the concentration of H_{2} and I_{2} (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)`