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# Quiz Solutions From 2/17/16 CH 102

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This 3 page Class Notes was uploaded by Jess Snider on Thursday February 18, 2016. The Class Notes belongs to CH 102 at University of Alabama - Tuscaloosa taught by Dr. Bakker in Winter 2016. Since its upload, it has received 54 views. For similar materials see General Chemistry in Chemistry at University of Alabama - Tuscaloosa.

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Date Created: 02/18/16

Quiz From 02/17/16 1. Given the following balanced equation, determine the rate of reaction with in terms of [NOCl]. 2 ???????? ???? + ???????? ???? 2→ 2 ????????????????(????) 1 ∆ ????????] ∆[????????2 1 ∆[????????????????] ???????????????? ???????? ???????????????????????????????? = − ∗2 ∆???? −→ − ∆???? −→ ∗2 ∆???? 1 ∆[????????????????] Answer: Rate=+ 2 ∆???? 2. Given the following balanced equation, determine the rate of reaction with respect 2o [????] . ???? ???? + 3 ???? ???? −→ 2 ???????? (????) 2 2 3 For any chemical equation of the form: aA + bB --------> cC + dD 1 ∆ ???? ] 1 ∆ ????[ ] 1 ∆ ????[ ] 1 ∆[????] ???????????????? = − ∗???? ∆???? ???????? − ???? ∆???? ????????????∗ ∆???? ???????? ????∗ ∆???? (Where t is the time at which the rate is measured) The negative signs shown in the equations (1) and (2) imply that the ∆[????] concentration of reactants are decreasing, which is making the te∆????sand d[B]/dt negative. Since the rate of a reaction is always positive, negative sign has been introduced in their rate equations. For the following reaction2 ???? ???? + 3 ????2???? −→ 2 ???????? (????3 1 ∆[2 ] Rate (with respect to2???? ) is given by: ????????????????3 ∆???? If the magnitude of change in the concentration is large, we can replace ∆[????2] ???????? ∆[???? 2 ????ℎ????????ℎ ????????????????????: ∆???? ∆???? Answer: ???????????????? = − ∗1 ∆[????2] 3 ∆???? 3. What is the expected freezing point of a 0.50 m solutio2 ????4 ???????? −→ 2 ???????? + + ????4 in water? ???? ???? for water is 1.8????° . Supposing complete ionization: 2???? ????4 −→ 2 ???????? + + ???????? (4-) (3 ions) (0.50???? ????????2???????? 4)∗ 3 = 1.50???? ???????????????? ???? (1.50???? ∗ (1.86° ) = 2.79°???? ????ℎ???????????????? ???? 0°???? − 2.79°???? = −2.8°???? Answer: -2.8˚C 4. What is the overall order of the following reaction, given the rate law? ???????? ???? + ???? ????3−→ ???????? ???? + ????2???? ) 2( ) ???????????????? = ????[????????][???? ]3 1???????? ???????????????????? ???????????? ???????? ???????????? 1???????? ???????????????????? ???????????? ???? =22???????? ???????????????????? ???????????????????????????? 1 + 1 = 2 Answer: 2 order 5. What are the units of k in the following rate law? ???????????????? = ???? ???? [????] 2 1 Answer: 2 ???? ???? 6. Determine the rate law and the value of k for the following reaction using the data provided. 2???????? ???? + ???? ???? 2→ 2 ???????? (????) 2 [????????] ????????) [????2 ????(????) Initial Rate (????????−1 ) −3 0.030 0.0055 8.55∗ 10 0.030 0.0110 1.71 ∗ 10−2 0.060 0.0055 3.42 ∗ 10−2 You must find the order of reaction for NO and O2 by using the chart. To find NO, use experiment 1 and 3 values. Divide both the rate and concentration (Exp 3 values÷Exp 1 values). Do the same process for O2Ho but with Exp 1 and Exp 2. If it is a 1:1 ratio, it is 1st order. If it is a 4:2 ratio, it is second order. If it is 1:0, it is zero order. To find K, pick one of the experiment values and use the order of reaction you found for each concentration then solve for k ???????????????? = ????[????] [????] ???? Answer: ???????????????? = 1.7 ∗ 10 ???? −2????−1 [????????] [???? ] 2 7. How many half-lives are required for the concentration of reactant to decrease to 12.5% of its original value? The definition of half-life is the time it takes for half the substance to decay 1 1 2 ???????? ????ℎ???? ????ℎ???????????? ???????? 2 1 1 1 2 ???????? 2 ????????4 1 1 1 2 ???????? 4 ???????? 8 ????ℎ????????ℎ ???????? 12.5% ???????? After the first half life, concentration of the reactant drops to 50%. After the second, it halves again to 25%, and after the third, 12.5%. Answer: 3 half-lives 8. The second-order decomposition of HI has a rate constant of 1.80 ∗ 10 ???? −1????−1 . How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M? 1 1 1 −3 1 [???? ???? = ???????? + [????] −→ [???? ???? = 1.80 ∗ 10 )(27.3 + 4.78 0 1 1 [???? ????= 0.258−→ ???? ???? =] 0.258 = 3.87???? Answer: 3.87 M −1 9. The first-order decomposition of ????2????at 1000K has a rate constant of 0.76???? . If the initial concentration of 2 ???? is 10.9 M, what is the concentration o2 ???? ????after 9.6 s? [????2????] ???? 0.76 ln( = −( )(9.6????) 10.9???? ???? [????2????] ???? −7.3???? 10.9???? = ???? −3 [????2???? = 7.4 ∗ 10 ???? Answer: 7.4 ∗ 10 −3???? 10. The second-order reaction 2 ????????(????????) −→ ???????? (????????) , has a rate constant equal to 3.0 ∗ 5 2 10 10 ???? −1????−1 at 25˚C. IF the initial concentration of ????????(????5) is 2.0 ∗ 10 ????, how long will it take for 90% of the reactant to disappear? Use integrated second order rate law, which has the general form [1]: ( ) = ???? ∗ ???? + ( 1 ) [????] [????0 (with [A] reactant concentration at time t, k rate constant, [A]₀ initial reactant concentration i.e. at t=0) For your reaction A is manganese pentacarbonyl Mn(CO)₅. Let X be the fraction of reactant which is consumed at time t, i.e. ∆ ????] ([????] −[????] [????] ???? = − = 0 = 1 − ( ) <=> [A] = [A]₀∙(1 - X) [????]0 [????0 [????0 Hence, 1 1 1 (([????] ∗ 1−???? ) = ???? ∗ ???? + [????]<=> (1−????) = [????]0∗ ???? ∗ ???? + 1 0 0 When solve this relation for t you get the time elapsed until a specified degree of conversion is reached: 1 ???? = 1−????−1 = ???? [????0 ∗???? ([????0 ∗????∗(1−????) For this problem: ???? ???? = ([????????(????????) ] ∗ ???? ∗ 1 − ???? ) ) 5 0 0.9 = (2.0 ∗ 10 ???? ∗ 3.0 ∗ 10 ????9 −1????−1 ∗ (1 − 0.9) −4 Answer: 1.5 ∗ 10 ????

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