The reaction 2 ClO2(aq) + 2 OH-(aq) ClO3-(aq) + ClO3-(aq) + H2O(l) was studied a

Question

The reaction 2 ClO2(aq) + 2 OH-(aq) ClO3-(aq) + ClO3-(aq) + H2O(l) was studied at a certain temperature with the following results:

Experiment [ClO2(aq)] (M) [OH-(aq)] (M) Rate (M/s)
1 0.0468 0.0468 0.0318
2 0.0468 0.0936 0.0636
3 0.0936 0.0468 0.127
4 0.0936 0.0936 0.254



(a) What is the rate law for this reaction?

Rate = k [ClO2(aq)] [OH-(aq)]
Rate = k [ClO2(aq)]2 [OH-(aq)]
Rate = k [ClO2(aq)] [OH-(aq)]2
Rate = k [ClO2(aq)]2 [OH-(aq)]2
Rate = k [ClO2(aq)] [OH-(aq)]3
Rate = k [ClO2(aq)]4 [OH-(aq)]





(b) What is the value of the rate constant?





(c) What is the reaction rate when the concentration of ClO2(aq) is 0.0613 M and that of OH-(aq) is 0.130 M if the temperature is the same as that used to obtain the data shown above?

M/s

Explanation / Answer

Here we go! First we need to determine the dependence of the rates on the reactant concentrations. [ClO2]0 [OH-]0 Initial Rate (mol/L) (mol/L) (mol L-1 s-1) 0.210 0.210 2.15 0.420 0.105 4.30 0.420 0.210 8.59 Sets 1 and 3 tell you that if you hold [OH-] constant and double [ClO2], the rate quadruples. That means there is a second-order dependence on [ClO2] (2^2 = 4). Sets 2 and 3 tell you that if you hold [ClO2] constant, and double [OH-], you double the rate of the reaction. That means there is a first-order dependence on [OH-]. Rate = k [OH-]*[ClO2]^2 Now we can assess the rate laws given for correctness: -d[OH-]/dt = k[OH-][ClO2-]2: Yes, this says that the rate of disappearance of OH- is first order in [OH-] and second order in [ClO2], as we just determined. d[ClO3-]/dt = k[OH-][ClO2]2: Yes, the appearance of ClO3- is first order in [OH-] and second order in [ClO2], as we just determined. -d[ClO2]/dt = k[ClO2]2[OH-]: Yes, this also says that the rate of disappearance of ClO2 is first order in [OH-] and second order in [ClO2], again as we just determined. d[ClO3-]/dt = k[ClO2][OH-]2: No, this is saying that the appearance of ClO3- is first order in [ClO2] and second order in [OH-]; we just found the opposite. d[ClO2-]/dt = k[OH-][ClO2]2: No, this says that the ***appearance*** of ClO2 is first order in [OH-] and second order in [ClO2]. This is true for the disappearance. ************************ #2: For Rate = -d[ClO2]/dt, calculate the rate constant (k). Here, we really could take any of the conditions above and solve for k; we'll just take the first one: [ClO2]0 [OH-]0 Initial Rate (mol/L) (mol/L) (mol L-1 s-1) 0.210 0.210 2.15 Rate = k [OH-]*[ClO2]^2; k = Rate / ([OH-]*[ClO2]^2) = (2.15 mol/L*s)/((0.210 mol/L)*(0.210 mol/L)^2 = (2.15 mol/L*s) / (0.210*0.210*0.210 mol^3/L^3) = 232 L^2/mol^2*s *********************** #3: What would be the initial rate (-d[ClO2]/dt) for an experiment with [ClO2]0 = 0.115 mol/L and [OH-]0 = 0.0758 mol/L? . Rate = k [OH-]*[ClO2]^2; we need the rate constant from #2: 232 L^2/mol^2*s Rate = k [OH-]*[ClO2]^2 = (232 L^2/mol^2*s) * (0.0758 mol/L) * (0.115 mol/L)^2 = 0.233 mol/L*s Hope that helped (and a big nod for your tenacity!)

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