Question

2) 20 points in the lab you studied the decomposition of peroxide in ACIDIC solution If the reaction is run under BASIC conditions a different result is obtained. 2 H2O2(aq) + 2 H2O(l) + O2(g) The rate of reaction can be determined by measuring the rate at which the volume of O2 increases at a pressure of latm. Use the data below to determine the rate law including! the value of the rate constant, “rate =k [H2021 [ITY". Show your work for full credit. PV = nRT R=0.08206 L atm/mol K hint: express the rate in A(mol O2)/At 0.1MI 0.9M H2O2 mL 02_produced OmL 2mL 4mL 6mL 8mL H20 run 1 10.00mL run 2 10.00mL run 3 20.00mL 10.00mL 5.00mL 15.00mL 10.00mL 5.00mL 0s O s 60s 31s 32s 110s 1625 219s 60s 92s 111s 673 885 115s

Answer 1

let the rate law of the reaction 2H20 I 2H₂0 (1) + O2(g) be nate = K[H2O₃] x [0279 – (i) but we have to find in terms of I gate = K [H202] * [1]8 - H₂O here is liquid so the rate laue will only contain the oxygen terms. And rate is the ratio of chang in concentration of oxygen to the tune Rate - Almol O2). Let the nolume be "v" A(At) VO Mote = concentration Volume Rate = (4-2) (110-60) Rate = viso - from ideal gas equation finding volume Rate = Vx50 PX V=nRT v= 1X 0.08 20 X 298 - 24.436 L

putting value of nolume in equation we get DI Rate 2 24.436450 - 1221.8 st i sort of sonst I] = 0.001636 -3 Rate = 1.636x10