Portage Learning Chem 104 Module 1 –Module 6 Exam Compiled

PORTAGE LEARNING CHEM 104 MODULE 1
–MODULE 6 EXAM Compiled



PortageLearningCHEM103Module1Exam: x, x, x, x, x, x,




Question 1 x,




Inthe reactionof gaseousN2O5 to yield NO2 gas andO2 gasasshown below,the followingdatatable is
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,




obtained:
x,




2N2O5 (g) → 4 NO
x,


+ O2 (g)
2 (g) x,
x,



x, x , x ,
x , x ,



x,




DataTable#2
x, x,




Time(sec) x,
[N2O5] [O2]

0 0.300 M x, 0

300 0.272 M x, 0.014 M x,




600 0.224 M x, 0.038 M x,




900 0.204 M x, 0.048 M x,




1200 0.186 M x, 0.057 M x,




1800 0.156 M x, 0.072 M x,




2400 0.134 M x, 0.083 M x,




3000 0.120 M x, 0.090 M x,




1. Usingthe [O2] data from the table, showthe calculation of the instantaneous rate earlyin the
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,

,reaction (0 secs to 300 sec).
x, x, x, x, x,




2. Usingthe [O2] data from the table, showthe calculation of the instantaneous rate latein thereaction (2400
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,




secs to 3000 secs).
x, x, x, x,




3. Explaintherelative values ofthe earlyinstantaneous rate andthe lateinstantaneous rate.
x, x, x, x, x, x, x, x, x, x, x, x, x,




YourAnswer:
x,




1. rate = (0.014 - 0) / (300 - 0) = 4.67 x 10-5 mol/Ls
x, x, x, x, x, x, x, x, x, x, x, x , x,




2. rate = (0.090 - 0.083) / (3000 - 2400) = 1.167 x 10-5 mol/Ls
x, x, x, x, x, x, x, x, x, x, x, x , x ,




3. Thelateinstantaneous rate issmallerthan theearlyinstantaneous rate.
x, x, x, x, x, x, x, x, x, x,

,Question 2 x,




Thefollowing rate data was obtained for thehypothetical reaction: A + B → X +
x, x, x, x, x, x, x, x, x, x , x , x , x , x , x , x , x , x , x , x , x , Y
Experiment# x, [A] [B] rate
1 0.50 0.50 2.0
2 1.00 0.50 8.0
3 1.00 1.00 64.0

1. Determinethe reaction orderwith respect to [A]. x, x, x, x, x, x, x,




2. Determinethe reaction orderwith respect to [B]. x, x, x, x, x, x, x,




3. Writetherate lawin theformrate= k [A]n [B]m (fillingin the correct exponents). x, x, x, x, x, x, x, x, x, x,
x , x,



x, x, x, x,




4. Showthecalculation of therate constant, k. x, x, x, x, x, x, x,




YourAnswer: x,




rate = k [A]x [B]y x, x, x,
x ,




rate1 / rate 2 = k [0.50]x [0.50]y / k [1.00]x [0.50]y
x, x, x, x, x, x, x,
x, x,



x, x,
x ,




2..0 =[0.50]x / [1.00]x
x, x, x, x,
x ,



x,




0.25 = 0.5x x, x,




x =2 x, x,




rate2 / rate 3 = k [1.00]x [0.50]y / k [1.00]x [1.00]y
x, x, x, x, x, x, x,
x, x,



x, x,
x ,




8..0 =[0.50]y / [1.00]y
x, x, x, x,
x,



x,




0.125 = 0.5y x, x,




y= 3 x, x,




rate = k [A]2 [B]3 x, x, x , x,




2.0 = k [0.50]2 [0.50]3 x , x, x , x,




k = 64
x, x,




Question 3 x,




ln [A] -ln [A]0 = - k t
x, x, x, x,


x ,
x, x, x, 0.693 = k t1/2 x, x, x,




An ancient sample of paper was found to contain 19.8 % 14C content as compared to a present-day sample.The
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,




t1/2 for14C is5720 yrs. Showthe calculationofthedecayconstant(k) andthe ageofthe paper.
x,


x,
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,

, YourAnswer: x,




0.693 = k t1/2 x, x, x,




0.693 = k (5720) x, x, x,




k = 1.21 x 10-4
x, x, x, x ,




ln [A] -ln [A]0 = - k t
x, x, x, x, x, x, x, x,




ln 19.8 - ln 100 = -1.21 x 10-4 t
x, x, x, x, x, x, x, x, x,
x ,




t = 13, 384 years
x, x, x, x,




Question 4 x,




Using the potential energy diagram below, state whether the reaction described by the diagram is
x, x, x, x, x, x, x, x, x, x, x, x, x, x,




endothermicorexothermicandspontaneousornonspontaneous,beingsuretoexplain youranswer.
x, x, x, x, x, x, x, x, x, x, x, x, x,




YourAnswer: x,




Thereactionisexothermicsinceithasanegativeheatofreactionanditisnonspontaneousbecauseit has
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,




relatively large Eact.
x, x, x,




Question 5 x,




Showthecalculationof Kcforthe followingreactionifan initialreactionmixtureof0.800moleof CO and 2.40
x, x, x, x,


x,
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,




mole of H2 in a 8.00 liter container forms an equilibrium mixture containing 0.309 mole of
x, x, x, x, x, x, x, x, x, x, x, x, x, x, x, x,




H2Oand corresponding amounts of CO, H2, and CH4.
x, x, x, x, x, x, x, x,




CO (g) x,



x , x ,
+ x , 3 H2 (g) x,



x, x , x ,
CH4 (g) + H2O(g) x, x ,
x , x , x,




YourAnswer: x,




0.309 mole of H2O formed = 0.309 mole of CH4 formed
x, x, x, x, x, x, x, x, x, x,