2 H2O2 (aq) <---> 2 H2O (l) + O2 (g) Rate of the reaction= - 1/2(d[H2O2]/dt) = (

Question

2 H2O2 (aq) <---> 2 H2O (l) + O2 (g)

Rate of the reaction= - 1/2(d[H2O2]/dt) = (d[O2]/dt) (in units of molarity)

Suppose you carry out the decomposition reaction using 25.0 mL of a 0.200 M hydrogen peroxide solution and obtain the following results: 8.26 mL of gas is collected in 120 seconds. Assume atmospheric pressure = 760 Torr (1 atm), room temperature = 25 degrees C and the water vapor pressure of water is 24 Torr. 2.73*10^-6 mol O2/s.

a. What is the rate of the reaction (d[O2]/dt)?

b. What is the change in the concentration of H2O2 per second (delta [H2O2]/dt)?

Explanation / Answer

a) gas obatined is Oxygen , gas vol = 8.26 ml , we have P = 760 torr = 1atm , T = 25+273=298 K, V = 8.26 ml = 0.00826 lites PV = nRT , n = PV/RT = 1x0.00826/0.0821 x298 = 0.000376 moles , time = 120 sec , d[O2]/dt = 0.000376/120 = 2.8 x10^ -6 M/s , b) -(1/2)d[H2O2]/dt = d[O2]/dt , d[H2O2]/dt = -2 d[O2]/dt = -2 x2.8 x10^ -6 = -5.6 x10^ -6 M/s

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