7. The oxidation of oxalate by permanganate (as shown below) was studied in aque

Question

7. The oxidation of oxalate by permanganate (as shown below) was studied in aqueous solution. I 6 H'(aq) + 5C,042-(aq) + 2 MnO4(aq) ? 2 Mn..(aq) 8H20 (I) + 10CO2(g) In one experiment, the concentration of permanganate ion was measured and recorded during the course of the reaction. The measurements allowed the experimenters to determine that at one point in the reaction, the permanganate ion (MnO4) was consumed at a rate 0.0173 M/s. At the same time, what is the rate of a. consumption of C20 ions? b. production of Mn2 ions? c. consumption of H ions? d. If the volume of the reaction solution is 225 mL, and the experiment is conducted at 25 °C and 1.00 atm what is the rate of production of CO2 in mL/min? The rate of consumption of MnO4 is still 0.0173 M/s, as before.

Explanation / Answer

ANSWER

16H+ + 5C2O42- + 2MnO4- -------> 2Mn2+ + 8H2O(l) + 10CO2(g)

Rate of reactio = Rate of formation of products = - rate of consumption of reactants

minus sign indicated reactants are consumed.

(a) 1/2 (rate of consumption of MnO4- ) = 1/5 (rate of consumption of C2O42- )

(rate of consumption of C2O42- ) = 5/2 (rate of consumption of MnO4- ) = 5/2 (0.0173M/s ) = 0.0432M/s

(b) 1/2 (rate of consumption of MnO4- ) = 1/2 (rate of consumption of Mn2+)

Or (rate of consumption of MnO4- ) = (rate of consumption of Mn2+) = 0.0713M/s

(c) 1/2 (rate of consumption of MnO4- ) = 1/16 (rate of consumption of H+)

(rate of consumption of H+) = 16/2 (rate of consumption of MnO4- ) = 16/2(0.0713M/s) = 0.1384M/s

(d)  1/2 (rate of consumption of MnO4- ) = 1/10 (rate of consumption of CO2)

(rate of consumption of CO2) = 10/2 ((rate of consumption of MnO4- ) = 5 X 0.0713M/s = 0.0865M/s

0.0865M/s = 0.0865M/s X 60 = 5.2M/min

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