## Questions

1 Complete the following table.

 Reaction equation Rate equation J = k ... Order Elementary or complex? H + Br2 → HBr + Br elementary S2O82- + 2I­- → 2SO42- + I2 [S2O82-] [I-] 1st order wrt to S2O82- 1st order wrt to I- 2nd order overall H2 + I2 → 2HI [H2][I2] CH3COCH3 + I2 → CH3COCH2I + H+ + I- reaction catalysed by acid H+ [CH3COCH3][ H+] H + Br2 → 2HBr 1st order wrt H2 order 0.5 wrt Br2

2 For an elementary reaction of the form A + A → products, write down the rate equation, find an expression for the concentration of A, cA, as a function of time and hence find an expression for the half-life of the reaction.

3 Write down expressions for the rate of the reaction 4Al + 3O2 → 2Al2O3 in terms of the changing concentrations of its reactants and its product.

4 The data in the table below refer to the reaction between phenolphthalein (‘Ph’) and excess OH- in solution

TABLE

(a) Show that the reaction is first order and determine its rate constant at the temperature at which the measurements were made.

(b) How might you determine the reaction order and rate constant from measurements of concentration in a reaction whose order was not already known?

5 Use the Arrhenius plot in Slide 10 to determine approximate values of A, Δε and ΔE for the reaction O + H2 → OH + H.

6 The pH in most cells is maintained by an equilibrium system (known as a ‘buffer’): H2PO4- + H2O <----> H3O+ + HPO42-

The acid H2PO4- has Ka = 6.3 x 10-8.

In a cell, total phosphate concentration [ HPO42-] + [H2PO4-] is typically about 2.0 mol dm-3 and [HPO42-] is close to 1.2 x 10-2 mol dm-3.

What is the cell pH?

7 Protein folding is an important process in biological physics (see Biological Molecules, Lecture 2):

protein (folded) <----> protein (unfolded)

The forward (unfolding) reaction of a certain protein (‘protein G’) has ΔH = +210.9 kJ mol-1 and ΔS = +616.7 J K-1 at 310 K.

(a) Calculate ΔG and the equilibrium constant K for the reaction at 310 K. Comment on your answer.

(b) How do ΔG and K change with increasing temperature?

1

 Reaction equation Rate equation J = k ... Order Elementary or complex? H + Br2 → HBr + Br [H][Br­2] 1st order wrt H 1st order wrt Br2 2nd order overall elementary S2O82- + 2I­- → 2SO42- + I2 [S2O82-] [I-] 1st order wrt to S2O82- 1st order wrt to I- 2nd order overall complex1 H2 + I2 → 2HI [H2][I2] 1st order wrt to H2 1st order wrt to I2 2nd order overall can’t tell2 CH3COCH3 + I2 → CH3COCH2I + H+ + I- reaction catalysed by acid H+ [CH3COCH3][ H+] 1st order wrt to CH3COCH3 zero order wrt I2 1st order wrt to catalyst H+ complex H + Br2 → 2HBr [H2][Br2]1/2 1st order wrt H2 order 0.5 wrt Br2 order 1.5 overall complex

1 If elementary, the rate here would be proportional to [I]2

2 Detailed studies show it to be complex, involving an intermediate dissociation of I­2 into I atoms.

2 J = - ½ dcA/dt = kcA2 (cf Equations 3 and 4: 2A → products)

1/cA(t) = 1/cA(0) + 2kt

t1/2 = 1/2kcA(0)

3 From Equation 3: J = - ¼ d[Al]/dt = - ⅓ d[O]/dt = ½ d[Al2O3]/dt

4 (a) A graph of ln[Ph} against t is a straight line, confirming that the reaction is first order (cf Equation 10). The magnitude of the gradient of the graph is k = 1.0 x 10-2 s-1.

(b) Plot c against t. Unless the graph is a straight line (zero order, which can probably be deduced from inspection of the data without the need to plot) draw tangents at several points to determine the rate, or use J(t) ≈Δc/Δt to find the average rate in each measured time interval. Plot log J against c and the order is equal to the gradient. Rate constant k can then be found using Equation 5.

5 From Equation 27, ln k = ln A -Δε/kBT.

Reading from the graph, gradient ≈ -5.7 /(1.3 x 10-3 K-1) = -4.4 x 103 K

Hence Δε = -4.4 x 103 K x 1.38 x 10-23 J K-1 ≈ 6.1 x 10-20 J.

Equation 21: ΔE = NA Δε = 6.02 x 1023 mol-1 x 6.1 x 10-20 J = 36 kJ mol-1.

From the graph, when 103/T = 1.2, ln k ≈ -30.3.

Substituting in Equation 27: ln (A/cm3 s-1) ≈ -25.1, A ≈ 1.3 x 10-11 cm3 s-1 (= 1.3 x 10-14 dm s-1)

6 From the data given: [HPO42-]/[H2PO4-] = 1.6; pKa = 7.2.

Equations 38 and 39: pKa = -log10{ [H3O+][HPO42-]/[H2PO4-] }

Equation 37: cell pH = - log10 [H3O+] = pKa + log101.6 = 7.4

7 (a) ΔG = 19.7 kJ mol, K = 4.75 x 10-4. At 310 K, only a very small fraction of protein G molecules are unfolded. This temperature (37 °C) is typical (human) cell temperature.

(b) At higher temperatures, ΔG becomes smaller and K increases i.e. a greater fraction of molecules that are unfolded. [Increasing the temperature by about 20 K approximately doubles the unfolded fraction.]