1. Kinetics – initial rates method

• In the reaction between bromate, bromide and hydrogen ion the following data were obtained; determine the rate law and value of the rate constant

Expt. [BrO3-](M) [Br-](M) [H+](M) Initial rate (mol/L.s)

1 0.10 0.10 0.09 8.0 x 10-4

2 0.192 0.10 0.10 1.6 x 10-3

3 0.20 0.20 0.09 3.2 x 10-3

4 0.102 0.10 0.20 3.18 x 10-3

• In the reaction between bromate, bromide and hydrogen ion the following data were obtained; determine the rate law and value of the rate constant

Expt. [BrO3-](M) [Br-](M) [H+](M) Initial rate (mol/L.s)

1 0.10 0.10 0.09 8.0 x 10-4

2 0.192 0.10 0.10 1.6 x 10-3

3 0.20 0.20 0.09 3.2 x 10-3

4 0.102 0.10 0.20 3.18 x 10-3

“By inspection”:

Expts 1 & 2 bromate concentration nearly doubles; rate doubles, so order w.r.t. bromate is 1.

Expts 2 & 3 bromide concentration doubles, rate doubles, so order w.r.t. bromide is 1.

Expts 3 and 4 bromate and bromide concentrations each fall by a factor of 2, so this contributes to the rate going down by a factor of 4. The hydronium ion concentration goes up by 2, but rate stays same, so order w.r.t. H+ is 2

Thus Rate = k [BrO3-] [Br-] [H+]2

Calculate rate constant by plugging the data for any experiment into the rate equation. k is about 8; what are the units?

2. Hess’s Law – simple example

The standard enthalpy for the hydrogenation of propene is -124 kJ mol-1.

The standard enthalpy for the combustion of propane is -2220 kJ mol-1.

The standard enthalpy for formation of water is -286 kJ mol-1.

Calculate the standard enthalpy of combustion of propene.

Comment on your answer.

a) The standard enthalpy for the hydrogenation of propene is -124 kJ mol-1.

b) The standard enthalpy for the combustion of propane is -2220 kJ mol-1.

c) The standard enthalpy for formation of water is -286 kJ mol-1.

Calculate the standard enthalpy of combustion of propene.

Comment on your answer.

a) C3H6 (g) + H2(g) C3H8(g) DHo = -124 kJb) C3H8 + 5O2 (g) 3CO2 (g) + 4H2O (g) DHo = -2220 kJc) H2 (g) + ½ O2 (g) H2O (l) DHo = -286 kJ

a + b - c gives an enthalpy change of -2058 kJ

Exothermic, as we’d expect for a combustion.

3. Entropy

When 1 mole of water is formed from the elements in their standard state, the absolute value of the enthalpy change is 286 kJ. Is this value positive or negative?

Calculate the entropy change in the surroundings when the reaction occurs.

How many litres of water does an Olympic-sized swimming pool hold? (Assume it’s 50m x 12m x 2m)

The standard entropy of vaporisation of water is 109 J K-1 mol-1. What is the enthalpy change when enough water to fill the pool condenses at 100oC?

Will the entropy of the surroundings increase or decrease as the water condenses?

Will the entropy of the surroundings increase or decrease as the water cools to room temperature?

When 1 mole of water is formed from the elements in their standard state, the absolute value of the enthalpy change is 286 kJ. Is this value positive or negative?

Negative (water is the result of burning hydrogen in oxygen).

Calculate the entropy change in the surroundings when the reaction occurs.

Use DS = qrev/T = DH/T = +286000/298 = 960 J K-1 mol-1

Comments:

Why is the DH term positive when obviously the reaction is exothermic?

Ans: because we are calculating the entropy change in the surroundings, not the system, and the surroundings are absorbing heat.

How many litres of water does an Olympic-sized swimming pool hold? (Assume it’s 50m x 12m x 2m)

1 m3 is 1000 litres, so it’s 50 x 12 x 2 x 1000 = 1,200,000 litres

The standard entropy of vaporisation of water is 109 J K-1 mol-1. What is the enthalpy change when enough water to fill the pool condenses at 100oC?

A mole of water weighs 18g and has an approximate volume of 18 mL. Therefore there are 55 moles water per litre, so the pool contains 66,000,000 moles of water.

The enthalpy change is 66,000,000 mol x 109 J = 7,194,000 kJ

Will the entropy of the surroundings increase or decrease as the water condenses?

Increase, since heat is released into the surroundings.

Will the entropy of the surroundings increase or decrease as the water cools to room temperature?

Ditto.

4. Electrochemistry

The standard reduction potentials of the Cu2+/Cu and Cu+/Cu couples are +0.34V and +0.522V respectively.

Calculate the standard emf of a cell in which Cu2+ and Cu+ are in equilibrium with solid copper.

Calculate the emf of the cell if the concentration of the Cu2+ solution is changed so that it is three times as large as that of Cu+.

What is DG for the new cell?

Is the reaction spontaneous or non-spontaneous?

The standard reduction potentials of the Cu2+/Cu and Cu+/Cu couples are +0.34V and +0.522V respectively.

Cu2+ (aq) + 2e- (aq) Cu (s) Eo = +0.34V Cu+ (aq) + e- (aq) Cu (s) Eo = +0.522V

Calculate the standard emf of a cell in which Cu2+ and Cu+ are in equilibrium with solid copper.

Suppose that the Cu2+ half-cell is on the right, then the emf of the cell (R-L) is +0.34 – 0.522 = -0.182V

Calculate the emf of the cell if the concentration of the Cu2+ solution is changed so that it is three times as large as that of Cu+.

Nernst equation is Ecell = Eo –RT/nF ln [Cu+]/[Cu2+] Eo we’ve just calculated. Put in the ratio of the concentrations of the two copper solutions and do the maths.

What is DG for the new cell?

DG = -nFEcell

Is the reaction spontaneous or non-spontaneous?

Depends on the sign of DG

5. Kinetics - units

At room temperature the rate constant for the reaction between oxygen atoms and ozone is 8.0 x 10-15 cm3 molecule-1 s-1.

In what units has the concentration of the two gases been measured?

Why is the rate constant so small?

What is the order of reaction?

A rate law has the form k [A]2[B]. What are the units of the rate constant if the reaction rate is measured in mol dm-3 s-1?

At room temperature the rate constant for the reaction between oxygen atoms and ozone is 8.0 x 10-15 cm3 molecule-1 s-1.

In what units has the concentration of the two gases been measured?

Must be molecule cm-3

Why is the rate constant so small?

Because in the rate constant we have inverse units of concentration (i.e., 1.0 upon molecules per cubic centimetre). The numerical value of the concentration will be extremely large, so k is inevitably small.

What is the order of reaction?

2nd order – the units of k tell us this.

A rate law has the form k [A]2[B]. What are the units of the rate constant if the reaction rate is measured in mol dm-3 s-1?

dm6 mol-2 s-1

6. Kinetics – half-lives

At 25oC the rate constant for the 1st-order decay of N2O5 is believed to be 3.38 x 10-5 s-1.

What is the half life of the reaction?

If the starting pressure is 500 mm Hg what will the pressure be after 250s? (Note that the integrated rate equation for a 1st order reaction is ln[A]t = -kt + ln[A]o)

If you had measured the gas pressure at several different times what should you plot to check whether your value for k was equal to the value quoted?

At 25oC the rate constant for the 1st-order decay of N2O5 is believed to be 3.38 x 10-5 s-1.

What is the half life of the reaction?

For a 1st order reaction t ½ = ln 2/k = 20507 s

If the starting pressure is 500 mm Hg what will the pressure be after 250s? (Note that the integrated rate equation for a 1st order reaction is ln[A]t = -kt + ln[A]o)

ln [A]t = -3.38 x 10-5 x 250 + ln 500

= - 0.00845 + 6.215 = 6.206 so [A]250 = 495.7 mm Hg

If you had measured the gas pressure at several different times what should you plot to check whether your value for k was equal to the value quoted?

ln [A]t vs t

7. Weak acid equilibria – straightforward example

Hypochlorous acid, HOCl is a weak acid with dissociation constant equal to 3.5 x 10-8.

Calculate the pH of a 0.1M solution of hypochlorous acid.

Hypochlorous acid, HOCl is a weak acid with dissociation constant equal to 3.5 x 10-8.

Calculate the pH of a 0.1M solution of hypochlorous acid.

[HOCl] [H+] [OCl-]

Initial 0.1 0 0

Change -x x x

Final 0.1-x x x

K = [H+] [OCl-] / [HOCl] = 3.5 x 10-8

Assume that x is small compared with 0.1, so that [HOCl]=0.1

We also know that the concentrations [H+] and [OCl-] are equal.

Thus [H+] = (3.5 x 10-9) ½ = 5.92 x 10-5 = [OCl-]

pH = -log10 [H+] = 4.2

8. Thermo

If water is well filtered to remove dust and then heated in a very smooth container it is possible to “super-heat” the water by several degrees above the boiling point. It will eventually boil suddenly and violently.

Calculate the entropy change of 1 mole of water and also of the surroundings when boiling occurs at 106oC. The enthalpy of vaporisation for water is 4070 J mol-1.

Comment on the result.

Note: the heat is released into the surroundings at a temperature of 106oC. For DS associated with the phase change we must use an equilibrium process; at what temperature does this process occur?

If water is well filtered to remove dust and then heated in a very smooth container it is possible to “super-heat” the water by several degrees above the boiling point. It will eventually boil suddenly and violently.

Calculate the entropy change of 1 mole of water and also of the surroundings when boiling occurs at 106oC. The enthalpy of vaporisation for water is 4070 J mol-1.

DS for the water is the change we’d get if the water boiled at 100oC (plus a small correction for warming the vapour, which we’ll ignore, since it is balanced by an equal and opposite change in the environment). So DS=DH/T=4070/373 = +10.91 J K-1 mol-1

For the surroundings DS = -4070/379 = -10.74 J K-1 mol-1

Comment on the result.

(i) DS positive for water because we are putting heat in; similarly, the surroundings are giving up heat, so entropy falls.

(ii) Overall gain in entropy as we’d expect.

Note: the heat is released into the surroundings at a temperature of 106oC. For DS associated with the phase change we must use an equilibrium process; at what temperature does this process occur?

373 K.

9. Hess’s law – more complicated

Calculate DH for the synthesis of diborane, B2H6, from its elements:

2B (s) + 3H2 (g) B2H6 (g)

2B(s) + 3/2 O2 (g) B2O3 (s) DH = -1273kJ

B2H6(g) + 3O2 (g) B2O3(s) + 3H2O (g) DH = -2035 kJ

H2 (g) + ½ O2 (g) H2O (l) DH = -286 kJ

H2O (l) H2O (g) DH = +44 kJ

Calculate DH for the synthesis of diborane, B2H6, from its elements:

2B (s) + 3H2 (g) B2H6 (g)

2B(s) + 3/2 O2 (g) B2O3 (s) DH = -1273kJ (a)

B2H6(g) + 3O2 (g) B2O3(s) + 3H2O (g) DH = -2035 kJ (b)

H2 (g) + ½ O2 (g) H2O (l) DH = -286 kJ (c)

H2O (l) H2O (g) DH = +44 kJ (d)

You should get (a) – (b) + 3 x (c) + 3 x (d), with a total enthalpy change of +36 kJ

10. Acid equilibria – more complicated

A solution contains 1.00M HCN, for which Ka = 6.2 x 10-10 and also 5.00M HNO2, for which Ka = 4.0 x 10-4.

Calculate (i) the pH of the solution and (ii) the CN- concentration in the solution at equilibrium.

A solution contains 1.00M HCN, for which Ka = 6.2 x 10-10 and also 5.00M HNO2, for which Ka = 4.0 x 10-4.

Calculate (i) the pH of the solution and (ii) the CN- concentration in the solution at equilibrium.

How do we deal with two acids? Ans: if one is much stronger than the other, ignore the contribution of the weak acid to the production of H+.

To calculate pH we do the normal ICE steps, leading to:

Ka = 4.0 x 10-4 = [H+][NO2-]/[HNO2] = x.x/(5.0-x) = x2/5, from which x=0.0447M.

X is the H+ concentration, so we can calculate pH=1.35

For the HCN, Ka=6.2 x 10-10 = [H+][CN-]/[HCN]

The concentration of the undissociated HCN is effectively 1.0, and we know that [H+]=0.0447M (ignore any H+ from the HCN, since it dissociates so little).

Thus [CN-]=6.2 x 10-10/0.0447 = 1.39 x 10-8M