Question

Chapter 14 and 15 1) Ozone decomposes to oxygen gas. 20, (g) ---> 30, (9) A proposed mechanism for this decomposition is o, =====> 0, +O (fast, equilibrium) O, + ---> 20% (slow) What is the rate law derived from this mechanism? 2) N2(g) + 3 H2(g) — > 2 NH3(g) Kc = 11.60 at 600 °C. Calculate Kp for this reaction at this temperature. 3) The rate of a particular reaction doubles when the temperature is increased from 250 C to 35° C. Calculate the activation energy for this reaction. 4) Hydrogen sulfide, a colorless gas with a foul odor, dissociates on heating: 10 pts 2H,S(g) - ------=> 2H, (g) + S, (g) When 0.100 mol H, S was placed in a 10.0 L vessel and heated to 1132°C, it gave an equilibrium Mixture containing 0.0285 mol H,. What is the value of K at this temperature. Set up the Data Table.

Answer 1

1) Guven - ozone decomposes into oxygengas 203 (g) - 302 (9) Proposed Mechanim O3 + O2 + 0 (fast, equilibrium) O3 +0 202 (slow) determined i We know by the that slow rate step of in a the reaction reaction. Hence in fact or equilibrium step- Rate of forward reach - Rate of backward reach het ki is the rate constant of the forward & Kal is the rate constant of the backward reach then Ki Cos] = ki [0] (02] - 0

② other species the top reach However in the now step- sate = ke [o] [03 - We need to replace (o] from because [o] is not a part of e. 203 (9) ► 30218). from equation - ki [03 ] = K [o] [02] [o] = ki [03]. K-1 [02] Put the value of [o] in equation ② rate = K2 Ki [03] [03] K-1 [02 ] het k is the final rate constant = kaki K-1

then rate = k [03] [03] [02] [03]² Irate = k You [02] [02] cone" increases rate of the reach decreases because O2 can react with o to form Oasten Reaction will proceed in the backward direction.