Chemical Equilibria Tutorial With Ans

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Chemical Equilibrium Raffles Junior College H2 Chemistry 2010/2011 Tutorial 7– Chemical Equilibria Mrs Jeannie Tan

1.

Methanol reacts with ethanedioic acid as shown: 2CH3OH (aq) + HOOC-COOH (aq)

CH3OOC-COOCH3 (aq) + 2H2O (l)

20 cm3 of 0.5 mol dm-3 of methanol was mixed with 30 cm3 of 0.4 mol dm-3 of ethanedioic acid. After the mixture has reached equilibrium, it was found that the 10 cm3 of the resultant mixture required 30 cm3 of 0.1 mol dm-3 NaOH for neutralization. Calculate the concentration of methanol at equilibrium and hence the value of Kc. [Ans: 0.02 mol dm-3, Kc = 1500] 2.

The key stage in the manufacture of sulphuric acid is the reaction between sulphur dioxide and oxygen. 2SO2 (g) + O2 (g) 2SO3 (g) (a) Draw a sketch graph showing how the rates of the forward and reverse reactions change from the time the two gases are mixed to the time the reaction reaches equilibrium. Label your two lines clearly. (b) Give an expression for Kp for this reaction, giving its units. (c) When a 2:1 mixture of SO2 and O2 are allowed to reach equilibrium at 500 °C and a total pressure of 5.0 atm, the partial pressure of SO3 was found to be 4.7 atm. Use your expression in (b) to calculate the value for Kp. [Ans : 5.52 x 103 atm-1]

3.

The equilibrium constant for the following reaction at 488 K is 2.00 x 10-6 atm: COCl2 (g) (a) (b)

Given that the total pressure of the system at equilibrium is PT and the degree of dissociation of COCl2 (g) is α, state the relationship between Kp, α and PT. Calculate the degree of dissociation, assuming temperature remains constant, at 1 atm pressure; [Ans: 1.41 x 10-3]

(Hint: when

4.

CO (g) + Cl2 (g)

is very small,

)

Steam can react with carbon monoxide under appropriate conditions according to the following reversible reaction: H2O (g) + CO (g) (a) (b) (c) (d)

CO2 (g) + H2 (g)

ΔH = - 40 kJ mol-1

Calculate amount of CO in the equilibrium mixture if 4 moles each of CO(g) and H2O(g) were placed in a vessel of constant volume at a temperature at which Kc is 9.0. What are the mole fractions of all the reactants and products. Sketch a graph to show how the amounts of the reactants and products change during the course of the reaction. Sketch 3 graphs showing only the amount of CO2 on the same scale to illustrate what happens when the reaction is repeated as before except that (i) the temperature is lowered,

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5.

Chemical Equilibrium

(ii) the volume of the vessel is decreased; and (iii) a catalyst is added to the reaction mixture with other factors remaining constant. Briefly describe what you have shown. [Ans: a) 1 mol, b) χH2 = χCO2 = 3/8, χH2O = χCO = 1/8] The partial oxidation of natural gases with steam involves an endothermic reaction as follows: CH4 (g) + H2O (g)

6*.

(a)

Give an expression for Kp for this reaction.

(a)

How is Kp and the composition of the equilibrium mixture affected by (i) increasing the temperature, (ii) decreasing the pressure, (iii) reducing the partial pressure of CO (g), (iv) adding an inert gas at constant pressure, (v) adding an inert gas at constant volume, (vi) reducing the volume of the reaction vessel, (vii) using a catalyst?

In their gaseous form, PH3 can combine with BCl3 to form a solid product PH3BCl3 at 80oC and all 3 species reached equilibrium in a flask. The equilibrium constant was determined to be 535 mol-2 dm6. (a) (b) (c) (d) (e) (f)

7*.

CO (g) + 3 H2 (g)

Write the reversible chemical equation of the above reaction. Calculate the concentration of PH3 and BCl3 in a closed vessel after some solid PH3BCl3 was formed at equilibrium. If the flask has a volume of 0.5 dm3, calculate the Kp for the above equilibrium. Some 0.005 mol of PH3 is now added into the flask. What is the new equilibrium concentration of BCl3 (g)? What happens when the solid PH3BCl3 is removed? Predict the effect on the equilibrium when it is heated. [Ans: b) (0.0432) c) (6.21 x 10-11 Pa-2) d) (0.0385)]

Phosphorus pentachloride dissociates into phosphorus trichloride and chlorine on heating as shown below: PCl5(g)

PCl3(g) + Cl2(g)

At 250 oC, it was found that 80% of 1 mole of PCl5 dissociated into PCl3 and Cl2. (a)

Express Kp in terms of total pressure, PT1, of the system.

[Ans: Kp = 1.78 PT1]

The pressure of the mixture was increased so that finally the mixture occupied half the original volume, with temperature remaining constant. (b)

What was the percentage dissociation at this new pressure, PT2, when the system had reached equilibrium? [Ans: 69.7 %]

(c)

What is the apparent molecular weight of PCl5(g) at this equilibrium? [Ans: 122.9]

Tutorial 7: Chemical Equilibria Suggested solutions 1.

Amt of CH3OH = 0.02 x 0.5 = 0.01 mol [CH3OH]mix = 0.01/0.05 = 0.2 mol dm-3

2

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Chemical Equilibrium

Amt of ethandioic acid = 0.03 x 0.4 = 0.012 mol [ethandioic acid]mix = 0.012/0.05 = 0.24 mol dm-3 Amt of NaOH = 0.03 x 0.1 = 0.003 mol (COOH)2 (aq) + 2NaOH (aq) → (COONa)2 (aq) + 2H2O (l) Amt of ethandioic acid reacted with NaOH = 0.5 x 0.003 = 0.0015 mol At eqm, [ethandioic acid] = 0.0015/0.01 = 0.15 mol dm-3 Initial conc/ mol dm-3 Change in conc Eqm conc/ mol dm-3

2CH3OH (aq) + HOOC-COOH (aq)

0.2 -0.18 0.02

0.24 -0.09 0.15

CH3OOC-COOCH3 (aq) + 2H2O (l)

0 +0.09 0.09

0 +0.18 0.18

Hence, Kc = [diester]/ [diacid] [alc]2= (0.09)/(0.15)(0.02)2 = 1500 mol-2 dm6 2.

(a)

Rate

forward

backward

time

3.

(b)

Kp = PSO32/ PSO22.PO2 atm-1

(c)

Eqm partial press of SO2 = 2/3 x (5.0 – 4.7) atm = 0.2 atm Eqm partial press of O2 = 1/3 x (5.0 – 4.7) atm = 0.1 atm Eqm partial press of SO3 = 4.7 atm Substituting into expression in (b), Kp = 5.52 x 103 atm-1

(a) Initial amt/mol Final amt/mol Mol fraction Partial press

COCl2 (g) c c(1-α) 1-α/1+α (1-α/1+α)PT

CO (g) + 0 cα α/1+α (α/1+α)PT

Cl2 (g) 0 cα α/1+α (α/1+α)PT

Kp = PCO. PCl2/ PCOCl2. Substituting,

(b) 4.

Subs PT = 1 atm, α = 1.41 x 10-3

(a) Initial amt/mol

H2O (g) + CO(g) 4 4

3

CO2 (g) + H2 (g) 0 0

Total = c(1+α)

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Chemical Equilibrium Eqm amt/mol

4-x

4-x

x

x

Let the volume of the vessel by V dm3. Kp = PCO2. PH2 / PH2O. PCO Substituting, (x/V)2/((4-x)/V)2 = 9 Solving, x = 3 mol Hence, eqm amt of CO = 1 mol (b)

(c)

(d)

Amt 4

Amt

(i) At lower temp

[CO2], [H2]

Original [CO2] curve (ii) Decreased vol (iii) With catalyst

3 2

[CO], [H2O]

1

Time

(d)

5.

6 *.

Time

(i)

At a lower temp, the rate at which eqm is attained is slower. By LCP, since ΔH < 0, the position of eqm shifts to the right and produce heat so as to increase the temp. Hence, amt of CO2 is higher at eqm. Kc is also increased.

(ii)

A decrease in volume, which leads to an increase in concentration, neither favours the backward nor forward reaction since no. of molecules on both sides of the reaction is the same. The amt of CO2 produced is the same as (c). A rise in pressure increases the rate at which equilibrium is attained.

(iii)

In the presence of a suitable catalyst, the rate at which eqm is attained is faster. However, there is no effect on the amt of CO2 formed. Kc remains the same. Catalyst increases the rate at which equilibrium is attained but has no effect on position of equilibrium.

(i)

Kp increases,

(ii)

Kp constant,

(iii)

Kp constant,

(iv)

Kp constant,

(v) (vi)

Kp constant, no change to partial pressures Kp constant,

(vii)

Kp constant, no change to partial pressures

(a) (b)

te

(a)

PH3 (g) + BCl3 (g)

PH3BCl3 (s)

(b)

Let y be the concentration of the reacting gases.

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Chemical Equilibrium Kc = [PH3]-1[BCl3]-1=

(c)

Amt of PH3 = amt of BCl3 = 0.0432 x 0.5 = 0.0216 mol

(d)

Conc of PH3 = 0.0432 + 0.005/0.5 = 0.0532 mol dm-3 (before eqm) Let z be the conc reduced. Kc = [(0.0532 – z)(0.0432 – z)]-1 = 535 Solving, z = 0.00468 or 0.0917 (rejected) Hence, [BCl3] = 0.0432 – 0.00468 = 0.0385 mol dm-3

(e)

Actually nothing visible will happen. If some solid is removed, the system is still at equilibrium as the concentration of PH3 and BCl3 remain unchanged. If ALL Solid is removed Actually nothing visible will happen. If the solid PH3BCl3 is removed, the system will no longer be at equilibrium (eqm). (A state of dynamic eqm requires the presence of all reactants and products) Position of eqm will shift to the right to produce some PH3BCl3. This will immediately lower both conc of PH3 and BCl3 and the position of eqm will shift to the left again until they become the initial eqm values. Thus essentially nothing seems to happen. The value of Kc is not defined.

(f)

7 *.

In the formation of PH3BCl3, a dative bond exists within the molecule of PH3BCl3. Thus the forward reaction is actually exothermic due to the formation of bonds. According to Le Chatelier’s Principle, when the system is heated, the position of eqm will shift to the left where heat is absorbed (because bond breaking will take place) and hence tends to reduce the temperature. At the same time, Kc will decrease.

(a) PCl5(g) 1.00 -0.80 0.20

Initial amt /mol Change in amt /mol Eqm amt /mol

(b)

----------- (1) Initial amt /mol Change in amt /mol Eqm amt /mol

PCl5(g) 1.00 -x 1-x

PCl3(g) + Cl2(g) 0 0 0.80 0.80 0.80 0.80

PCl3(g) + Cl2(g) 0 0 x x x x

----------- (2) 5

Total no = 1.80 mol

Total no = 1 + x

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Chemical Equilibrium

Assuming ideal gas behaviour, PV = nRT

----------- (3)

Substituting (3) into (2),

----------- (4) At constant temperature, Kp remains unchanged. Hence equating (1) and (4),

(c)

Solving, x = 0.697 (3 sf)

Hence percentage dissociation = 69.7%

Apparent molecular mass

=

= Substituting

x

Mr(PCl5) +

(208.5) +

Mr(PCl3) +

(137.5) +

(71.0)

= 0.697. Thus, apparent molecular mass

OR Apparent molecular mass = Hence, apparent molecular mass

= 122.9

6

Mr(Cl2)

= 122.9