Chem 146 Problem Set 2 Rate-k[DJ [E\' For questions 1-5 refer to expression e1.

Question

Chem 146 Problem Set 2 Rate-k[DJ [E' For questions 1-5 refer to expression e1. 1. The rate of reaction is order overall. 2. If rate has units ofmolesML sec, than k has units of 3. At a given temperature if the concentration of A is tripled, while the concentration of Band Care held constant, then the rate of reaction wi (increase or decrease pick one) by a factor of 4. At a given temperature ifthe concentration of Bis tripled, while the concentration of A and Care held constant, then the rate of reaction will (increase or decrease-pick one) by a factor of 5. At a given temperature if the concentration of C is tripled, while the concentration of A and Bare held constant, then the rate of reaction will increase or decrease pick one) by a factor of For questions 6-10 refer to expression #2. 6. The rate of reaction is order overall. 7. If rate has units of moles/L sec, than k has units of 8. At a given temperature ifthe concentration of Dis tripled, while the concentration of Eis held constant, then the rate of reaction will increase or decrease pick one) by a factor of 9. At a given temperature if the concentration of E is tripled, while the concentration of Disheld constant, then the rate of reaction will (increase or decrease-pick one) by a factor of 10. At a given temperature if the concentration ofD and Eare doubled, then the rate of reaction will (increase or decrease pick one by a factor of

Explanation / Answer

. The rate is given by Rate= K[A]2 [B]0.5 [C]-0.5, The overall order is sum of powers of concentration tems in the rate expression. Hence order= 2+0.5-0.5= 2

2.Concentration is measured in moles/L

Hence rate= Moles/l.sec = K* (Moles/L)2

K=L/Mole.sec

3. concentration of A is tripled, new concentration of A= 3[A], while keepng [B] and [C] constant

New Rate = K[3A]2 [B] 0.5[C]0.5 = 9 [A]2 [[B] 0.5[C]-0.5 = 9* initial rate . The rate increases by a factor of 9.

4. when the concentrationof B is tripled ( while keeping [A] and [C] constant

, the new rate = K[A]2 [3B] 0.5[C]0.5= 1.732[A]2 [B]0.5[C]-0.5 =1.732 times initial rate. The rate increases by a factor of 1.732

5. When concentrationo of C is tripled while keeping [A] and [B] constnat,

New rate = K[A]2 [B]0.5 [3C]-0.5 = 0.577* inital rate. The rate is reduced by 0.577

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