Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 7.17:

K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

Answer:

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.

Now, according to the reaction,

We can write,

Hence, at equilibrium,

Question 7.18:

Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

(i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Answer:

(i) Reaction quotient, 

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

Therefore, equilibrium constant for the given reaction is:

(iii) Let the volume of the reaction mixture be V.

Therefore, the reaction quotient is,

Since , equilibrium has not been reached.

Question 7.19:

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10^–1 mol L^–1. If value of Kc is 8.3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

Answer:

Let the concentrations of both PCl₃ and Cl₂ at equilibrium be x molL^–1. The given reaction is:

Now we can write the expression for equilibrium as:

Therefore, at equilibrium,

Question 7.20:

One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.

FeO (s) + CO (g)   Fe (s) + CO₂ (g); Kp= 0.265 at 1050 K.

What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: p_CO = 1.4 atm and  = 0.80 atm?

Answer:

For the given reaction,

Since , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO₂ will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO₂ be p.

Then, we can write,

Therefore, equilibrium partial of 

And, equilibrium partial pressure of 

Question 7.21:

Equilibrium constant, Kc for the reaction

at 500 K is 0.061.

At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L^–1 N₂, 2.0 mol L^–1 H₂ and 0.5 mol L^–1 NH₃. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

Answer:

The given reaction is:

Now, we know that,

Since , the reaction is not at equilibrium.

Since  , the reaction will proceed in the forward direction to reach equilibrium.

Question 7.22:

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:

for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10^–3 molL^–1, what is its molar concentration in the mixture at equilibrium?

Answer:

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

Now, we can write,

Therefore, at equilibrium,

Question 7.23:

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass

Calculate Kc for this reaction at the above temperature.

Answer:

Let the total mass of the gaseous mixture be 100 g.

Mass of CO = 90.55 g

And, mass of CO₂ = (100 – 90.55) = 9.45 g

Now, number of moles of CO, 

Number of moles of CO­₂, 

Partial pressure of CO,

Partial pressure of CO₂,

For the given reaction,

Δn = 2 – 1 = 1

We know that,

Question 7.24:

Calculate a) ΔG°and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

where ΔfG° (NO₂) = 52.0 kJ/mol

ΔfG° (NO) = 87.0 kJ/mol

ΔfG° (O₂) = 0 kJ/mol

Answer:

(a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 + 0}

= – 35.0 kJ mol^–1

(b) We know that,

ΔG° = RT log Kc

ΔG° = 2.303 RT log Kc

Hence, the equilibrium constant for the given reaction K_c is 1.36 × 10⁶