MC test: The equilibrium law

Multiple choice test on 17.1 The equilibrium law

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

2 mol of a gaseous compound P were placed in a vessel. Compound P partially decomposed when heated to produce two gaseous compounds, Q and R.
At equilibrium, x mol of R were present and the total number of moles present was (2 + x).

Which is a possible equation for this equilibrium reaction?

At equilibrium, amount of P + amount of Q + amount of R = (2 + x) mol.
But amount of R = x mol, so at equilibrium, amount of (P + Q) = 2 mol.
Since the initial amount of P = 2 mol and the initial moles of Q = 0 mol, the total amount of P and Q does not change during the reaction. Hence the stoichiometric coefficients of P and Q are equal. An actual reaction that would fit this is PCl5(g) ⇄ PCl3(g) + Cl2(g) but other equations (which are not in the list above) such as 2P(g) ⇄ 2Q(g) + R(g) would also fit the data. 

 

1.50 mol of A and 1.00 mol of B  were placed in a 5.0 dm3 reaction flask. After equilibrium has been established, at a temperature of 450 K, 0.50 mol of C were present.

The stoichiometric equation for the reaction is: A(g) + B(g) ⇌ C(g) + D(g)

What is the equilibrium constant for the reaction at this temperature?

At equilibrium: Amount of C = amount of D = 0.50 mol; [C] = [D] = 0.5/5.0 = 0.1 mol dm-3. Amount of A = 1.50 − 0.5 mol so [A] = 1.0/5.0 = 0.20 mol dm-3 and amount of B = 1.00 − 0.5 mol so [B] = 0.50/5.0 = 0.10 mol dm-3.
Kc = [C][D] / [A][B] = 0.12 ÷ (0.20 x 0.10) = 0.50

 

According to a research paper written by A.H. Taylor Jn. and R.H. Crist in the Journal of the American Chemical Society in 1941, hydrogen iodide is 22.3% dissociated into hydrogen and iodine at 730.8 K.

Which expression gives the value of Kc at 730.8 K for 2HI(g) ⇄ H2(g) + I2(g)

Assume there is initially 1 mole of HI. The amount of HI remaining at equilibrium = (1 − 0.223) = 0.777 mol. Each mol of HI produces 0.5 mol of H2 and 0.5 mol of I2 when it dissociates so  H2 at equilibrium = amount of I2 at equilibrium = 0.5 x 0.223 = 0.1115 mol.  If the total volume is V then [HI] = 0.777/V and [H2] = [I2] = 0.1115/V. Kc = [H2] [I2] / [HI]2 = (0.1115/V)2 ÷ (0.7777/V)2 = 0.11152 ÷ 0.7772.

 

Phosphorous pentachloride can be formed by the reaction between chlorine and phosphorous trichloride.

Cl​2(g)​ + PCl​3(g)​ ⇄ PCl​5(g) ​ ​

0.20 mol of chlorine and 0.50 mol of phosphorus trichloride​ were placed in a 100 cm3 flask and the reaction allowed to reach equilibrium at a fixed temperature.  At ​equilibrium, 0.14 mol of phosphorus pentachloride​ was produced.

Which expression gives the value of Kc for this reaction at the temperature of the reaction?

Equilibrium amounts: Cl2 : (0.20 − 0.14); PCl3 : (0.50 − 0.14); PCl5 : 0.14 mol
Equilibrium concentrations: Cl2 : 0.06/0.100; PCl3 : 0.36/0.100; PCl5 : 0.14/0.100 mol dm−3
Kc = [PCl5]/[Cl2][PCl3] = 1.4 ÷ (0.6 x 3.6)

 

1.0 mol of ethanol and 2.0 mol of ethanoic acid was reacted together in the presence of a small amount of acid catalyst and the mixture left to reach equilibrium at 293 K. At equilibrium 0.845 mol of ethyl ethanoate was found to be present.

H3CCOOH(l) + C2H5OH(l) ⇄ H3CCOOC2H5(l) + H2O(l)

Which expression gives the value of Kc for this esterification reaction at 293 K?

Equilibrium amounts: H3CCOOH : (2.0 − 0.845); C2H5OH : (1 − 0.845); H3CCOOC2H5 : 0.845; H2O : 0.845 mol
Equilibrium concentrations (in volume V dm3): H3CCOOH : 1.155/V; C2H5OH : 0.155/V; H3CCOOC2H5 = H2O = 0.845/V mol dm−3
Kc = ([ester] x [water]) ÷ ([acid] x [alcohol]) = 0.8452 ÷ (1.155 x 0.155). The volume is not necessary as it cancels out because there are the same number of moles on both sides in the stoichiometric equation.

 

A mixture consisting of 0.60 mol of ethanoic acid, 0.40 mol of ethanol, 0.20 mol of ethyl ethanoate and 0.60 mol of ​water was placed in a 250 cm3 sealed flask and left to reach equilibrium at 293 K.  ​At equilibrium it was found that the mixture contained 0.40 mol of the ester.

What is the value of Kc at 293 K for the hydrolysis reaction: H3CCOOC2H5(l) + H2O(l) ⇄ H3CCOOH(l) + C2H5OH(l) ?

Since the amount of ester has increased by 0.20 mol the amount of water will also have increased by 0.2 mol and the amount of acid and alcohol will each have decreased by 0.20 mol.
Equilibrium concentrations: H3CCOOH : 0.40/0.250; C2H5OH : 0.20/0.250; H3CCOOC2H5 : 0.40/0.250; H2O : 0.80/0.250 mol dm−3
Kc = ([acid] x [alcohol]) ÷ ([ester] x [water])  = (0.40 x 0.20)  ÷ (0.40 x 0.80) = 0.25

 

The equilibrium constant for the reaction between sulfur dioxide and oxygen is 680 at 373 K.

2SO2(g) + O2(g) ⇄ 2SO3(g)

At equilibrium at 373 K it is found that 0.600 mol of oxygen and 0.600 mol of sulfur dioxide are present in a 4.00 dm3 flask which only contains oxygen, sulfur dioxide and sulfur trioxide.

Which expression will give the concentration of sulfur trioxide in the flask?

[O2] = 0.600/4 mol dm−3 and [SO2] = 0.600/4 mol dm−3. Kc = [SO3]2 ÷ ([O2][SO2]2)
[SO3]2 = 680 x (0.0.150)3 so [SO3] = (680 x 0.1503)1/2 mol dm−3 

 

What does K stand for in the expression ΔG⦵ = − RT lnK ?

ΔG⦵ = − RT lnK is the classic expression relating equilibrium to Gibbs free energy.

 

Water dissociates to form hydrogen ions and hydroxide ions.

H2O(l) ⇄ H+(aq) + OH−(aq) ΔH⦵ = + 56 kJ mol−1; ΔS⦵ = − 81 J K−1 mol−1

Which expression gives the value for the ionic product of water, Kw at 298 K?

(R = 8.31 J K−1 mol−1)

ΔG⦵ = − RT lnKw = ΔH⦵ − TΔS⦵ 
lnKw = − (ΔH⦵ − TΔS⦵)/RT
Kw = = e− ((56000 − (298 x − 81)) ÷ (8.31 x 298))

 

Which row gives the correct combination for the values of Kc and ΔG⦵ for the stated equilibrium mixture?

RowEquilibrium mixtureKcΔG⦵
1Contains mainly products>> 1positive
2Contains mainly reactants>> 1negative
3Contains mainly reactants<< 1positive
4Contains mainly products<< 1negative

When Kc << 1 it will contain mainly reactants and ln Kc will be negative. From ΔG = − RT lnKc it can be seen when ln Kc is negative ΔG will be positive.

 

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