Questions
1 Complete the following table.
Reaction equation 
Rate equation J = k ... 
Order 
Elementary or complex? 
H + Br_{2} → HBr + Br 
elementary 

S_{2}O_{8}^{2} + 2I^{} → 2SO_{4}^{2} + I_{2} 
[S_{2}O_{8}^{2}] [I^{}] 
1^{st} order wrt to S_{2}O_{8}^{2} 1^{st} order wrt to I^{} 2^{nd} order overall 

H_{2} + I_{2} → 2HI 
[H_{2}][I_{2}] 

CH_{3}COCH_{3} + I_{2} → CH_{3}COCH_{2}I + H^{+} + I^{} reaction catalysed by acid H^{+} 
[CH_{3}COCH_{3}][ H^{+}] 

H + Br_{2} → 2HBr 
1^{st} order wrt H_{2} order 0.5 wrt Br_{2} 
2 For an elementary reaction of the form A + A → products, write down the rate equation, find an expression for the concentration of A, c_{A}, as a function of time and hence find an expression for the halflife of the reaction.
3 Write down expressions for the rate of the reaction 4Al + 3O_{2} → 2Al_{2}O_{3} 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 + H_{2} → OH + H.
6 The pH in most cells is maintained by an equilibrium system (known as a ‘buffer’): H_{2}PO_{4}^{} + H_{2}O <> H_{3}O^{+} + HPO_{4}^{2}
The acid H_{2}PO_{4}^{} has K_{a} = 6.3 x 10^{8}.
In a cell, total phosphate concentration [ HPO_{4}^{2}] + [H_{2}PO_{4}^{}] is typically about 2.0 mol dm^{3} and [HPO_{4}^{2}] 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?
Answers
1
Reaction equation 
Rate equation J = k ... 
Order 
Elementary or complex? 
H + Br_{2} → HBr + Br 
[H][Br_{2}] 
1^{st} order wrt H 1^{st} order wrt Br_{2} 2^{nd} order overall 
elementary 
S_{2}O_{8}^{2} + 2I^{} → 2SO_{4}^{2} + I_{2} 
[S_{2}O_{8}^{2}] [I^{}] 
1^{st} order wrt to S_{2}O_{8}^{2} 1^{st} order wrt to I^{} 2^{nd} order overall 
complex^{1} 
H_{2} + I_{2} → 2HI 
[H_{2}][I_{2}] 
1^{st} order wrt to H_{2} 1^{st} order wrt to I_{2} 2^{nd} order overall 
can’t tell^{2} 
CH_{3}COCH_{3} + I_{2} → CH_{3}COCH_{2}I + H^{+} + I^{} reaction catalysed by acid H^{+} 
[CH_{3}COCH_{3}][ H^{+}] 
1^{st} order wrt to CH_{3}COCH_{3} zero order wrt I_{2} 1^{st} order wrt to catalyst H^{+} 
complex 
H + Br_{2} → 2HBr 
[H_{2}][Br_{2}]^{1/2} 
1^{st} order wrt H_{2} order 0.5 wrt Br_{2} 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 I^{}_{2} into I atoms.
2 J =  ½ dc_{A}/dt = kc_{A}^{2} (cf Equations 3 and 4: 2A → products)
1/c_{A}(t) = 1/c_{A}(0) + 2kt
t_{1/2} = 1/2kc_{A}(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/k_{B}T.
Reading from the graph, gradient ≈ 5.7 /(1.3 x 10^{3} K^{1}) = 4.4 x 10^{3} K
Hence Δε = 4.4 x 10^{3} K x 1.38 x 10^{23} J K^{1} ≈ 6.1 x 10^{20} J.
Equation 21: ΔE = N_{A} Δε = 6.02 x 10^{23} mol^{1} x 6.1 x 10^{20} J = 36 kJ mol^{1}.
From the graph, when 10^{3}/T = 1.2, ln k ≈ 30.3.
Substituting in Equation 27: ln (A/cm^{3} s^{1}) ≈ 25.1, A ≈ 1.3 x 10^{11} cm^{3} s^{1} (= 1.3 x 10^{14} dm s^{1})
6 From the data given: [HPO_{4}^{2}]/[H_{2}PO_{4}^{}] = 1.6; pK_{a} = 7.2.
Equations 38 and 39: pK_{a} = log_{10}{ [H_{3}O^{+}][HPO_{4}^{2}]/[H_{2}PO_{4}^{}] }
Equation 37: cell pH =  log_{10} [H_{3}O^{+}] = pK_{a} + log_{10}1.6 = 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.]