Question

Experiment: Reaction Kinetics Aim: To find the reaction orders and rate constants for the following reaction: 3 H2C2O4(aq) + 2 MnO4 (aq) --> 6 CO2+ 2 OH(aq) + 2 H2O + 2 MnO2s In other words to obtain m, n, and k for Rate = k[MnO4"[H2C,0.)" Trial 1 5 Trial 2 | 10 Trial 3 5 Volume of Oxalic Acid (mL Volume of Permanganate (mL) Volume of Water (mL) Results: Average Time (seconds) 343.7 Concentration Oxalic Acid Concentration Rate of Permanganate Reaction Trial 1 Trial 2 Trial 3 156.3

1. Determine which reactant is limiting the rate of reaction. Justify the answer.
2. Determine the rate of reaction in trial 1,2,3
3. What are the experimental values for n and m?

Thank you so so much for your help.

Answer 1

At first, we have to find the concentration of each reactant in all three trials, then by the inverse of t average we will find k for all three trials and so on calculation is shown in given picture.
Page Trial o concentration of [Mnoy J = 0.130 X Vuscel - Viorel Votal I St1+6 = 12 Vused - Iml = 0.01084 . [Mnoy J -Terial ② @ concentration of = 0.130x L mamou J - 12 - 0.1308 Vusal Vuocd = 1 ml Torel 12 . VTotal = same as im Trial 1 = 0.01084 Thial ③ o Concentration (Mnoy I = 0.1308 Vused Totul Kosed = 2 => 0.130 X 2 12 = 0.02167 Trial o concentration of oxalic Acid = 0.755x 5 -0.315 2 Same as Trials = 0.315 Trial Concentration of oxalic Acid = 0.755x10 - 0 6292 12

classmate Torial o tau = 343.7 K =- = K = 0.0029095 Slami lovely y = 976 = 0.0002459 - T563 = 0.006397 average K = Kitkat Kz 1 Rase, k[ mnou] CH GOUT Rate Kem noi mau CO u In 0.0065174667 (0.315 0.629 Rare - Kenmou) [12C2O4], KEM nou I [h CGO 4 J 1 = 0.02767 Ratez (0.01084" 10.02167) m obog (0.315). m m-m can solving all three Wenget n = 1 m=0

classmate Date Page S Rade of Reactions 0 R,= Kat 0.315 ] 0,002 9095 x 0. 315 0.000916 K, 10.62920 = 0,0102459 x 0.6292 - 0.0 06446 - R same as R, nearly and limitry Rate's. Meactant Cmnou as in all telial it's concentration is lesser in Quantity