Question

Bromate and bromide react in acidic solution.

                        BrO₃^-(aq) + 5 Br^-(aq) + 6 H⁺(aq) -> Br₂(aq) + 3 H₂O(l)

The kinetics were studied by measuring the rate of appearacne of Br₂ (orange color in solution). Use the following data to determine the rate law (i.e. determine E_a, x, y, and z).

                                    rate = Ae^-Ea/RT [BrO₃^-]^x[Br^-]^y[H⁺]^z

            KBr                 KBrO₃             H⁺                   H₂O                 T         initial rate

            0.0020M         0.0020M         0.020M

            ---------------------------------------------------------------------

            10.0mL            10.0mL            10.0mL            70.0mL            27^oC    1.44 x 10^-12M/s

            20.0mL            10.0mL            10.0mL            60.0mL            27^oC    2.88 x 10^-12M/s

            10.0mL            20.0mL            10.0mL            60.0mL            27^oC    2.88 x 10^-12M/s

            10.0mL            10.0mL            20.0mL            60.0mL            27^oC    5.76 x 10^-12M/s

            10.0mL            10.0mL            10.0mL            70.0mL            43^oC    4.63 x 10^-12M/s

Answer 1

Part A, To determine order of the reaction.

Look at the first and second run,

when we double the concentration of KBr and keep the other two constant the rate doubled, so the rate is first order with respect to [Br-] concentration.

Look at the first and third run,

when we double the concentration of KBrO3 while keeping the other two the same, the rate doubled, so the reaction is first order with respect to [BrO3-] concentration.

Look at the first and the fourth run,

when we double the concentration of H+ while keeping the other two reactant concentration the same, the rate quadrupled, so the rate is second order with respect to [H+] concentration.

So we can write for this reaction, the rate can be represented as,

rate = k [BrO3-][Br-][H+]^2

where,

k = rate constant.

Thus,

x = 1

y = 1

z = 2

Part B : To determine Ea for the reaction.

We would use Arrhenius law to calcuate Ea,

ln(k2/k1) = Ea/R[1/T1-1/T2]

Take the first run and the last run readings for this calculation.

We have,

k1 = 1.44 x 10^-12 M/s

k2 = 4.63 x 10^-12 M/s

T1 = 27 oC = 27 + 273 = 300 K

T2 = 43 oC = 43 + 273 = 316 K

R = 8.314 J/mol.K

Feed the values we get,

ln(4.63 x 10^-12/1.44 x 10^-12) = Ea/8.314 [1/300 - 1/316]

Ea = 57532.697 J/mol = 57.54 kJ/mol