QUIZ#3                                               CHEM100.11     February 15, 2012

 

                                                                         Name: ___________________

                                                                                    Family name    First name

 

PLEASE READ THIS BEFORE YOU BEGIN:  This test comprises 10 pages (including this page) and a Periodic Table.  Please make sure you have a complete exam and that you have filled in the information above.  All questions are to be answered on the question sheets.   Calculators are permitted.  You have 50 min. 

 

Useful equations, constants and conversion factors:

R = 8.314 J/mol K                       

J = kg m2 s-2

1 cal = 4.18 J

 


Arrhenius equation:                       Integrated forms:  

 


    

differential rate law: rate = k [A]n [B]m [C]p…..

Integrated Rate Laws:            ln[A]t  =  -kt   +  ln[A]0

 

 

 

 

 

 

 


qsolution = -q reaction = m × specific heat × DT

 

DG°  =  DH°   -   TDS°

DG  = DG°  + RTlnQ              DG°  = - RT lnKeq

DHrxn = ΣnDHf0products - ΣnDHf0reactants

 

 

SHORT ANSWERS {4problems x10 scores, 40 scores, }

  1. {10} A first-order reaction has a rate constant of 2.75x10-2 s-1 at 20.0 oC. 

(6)   What is the value of k at 60.0 oC if Ea = 75.5 kJ/mol?

{4} Calculate half-life for each temperature.

 

2. {10} A pound (454g) of plain M&M candies contains 96 g of fat, 320 g of carbohydrate, and 21 g of protein.  What is the energy value, in dietary (nutrition) Calories, of this amount of candies?

Fuel values:

carbohydrate and  protein :   17 kJ/g

fat                                              38 kJ/g

 

 

3. {10} Consider the reaction

3CH4(g)  D  C3H8 (g) + 2H2 (g)

For this rxn, ΔGo = 128.9 kJ/mol                      

Calculate ΔG at 298 K for the rxn mixture that consists of 40.0 atm CH4 , 0.01 atm C3H8  and 0.018 atm H2  

 

 

  1. {10} SO2Cl2 decomposes in the gas phase by the reaction

SO2Cl2 (g) à SO2 (g) + Cl2 (g)

The reaction is first order in SO2Cl2 and the rate constant is 3.0 x 10-6 s-1 at 600 K. A vessel is charged with 2.4 atm of SO2Cl2 at 600K. Calculate the partial pressure of SO2Cl2 after 3.0 x 105 s.

 

Longer Problems {3 problems x 20 scores}

  1. {20} When a 4.25-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 22.0 oC to 16.9 0C. Calculate  ΔH (in kJ/mol NH4NO3) for the solution process:

Assume that specific heat of the solution in calorimeter is the same as for water, 4.184 J/g oC.

 

6. {20}

a.      {15} Using the date from the Table below, calculate the equilibrium constant at 100 oC for the following rxn:

2SO2 (g) + O2 (g) D 2 SO3 (g)

b.      {4} At which temperature reaction will be at equilibrium?

c.       {1} Would you increase or decrease temperature in order to make the reverse rxn spontaneous?

Table

substance                    ΔHof (kJ/mol)                         S (J/mol K)

SO2 (g)                        -297                                         249

O2 (g)                          0                                              205

SO3 (g)                        -395                                         256

 

7. {20} the data below were obtained for the rxn:

aA + bB + cC à dD + eE

Exp.       [A], M          [B], M             [C]                   Initial rate, M/s

1          0.020               0.010               0.01                 1.15 x 10-2

2          0.040               0.010               0.01                 1.15 x 10-2

3          0.030               0.020               0.01                 2.30 x 10-2

4          0.020               0.020               0.005               5.75 x 10-3

(a)   {15} write down the rate law for this rxn

(b)   {1} what is the overall order for this rxn?

(c)    {3} calculate the rate constant for this rxn (show units!)

(d)   (1) what is the initial rate at [A] =[B] = [C] =0.1? 

 

Answers:

  1. ln(k1/k2) = 75500/8.314 (1/293-1/333)= 3.723; k1/k2= 41.4;  k1(60)=0.0275x41.4=  1.13 s-1

t 1/2 (20oC) = 0.693/0.0275 s-1 = 25.2 s;   t 1/2 (60 oC) = 0.693/1.13 s-1 = 0.613 s

 

  1. 96g x 38 kJ/mol + (320 g + 21 g) 17 kJ/g = 9445 kJ

2249                    kcal or 2249 Cal

 

  1. DG  = DG°  + RTlnQ ;

Q = P(propane)P2(H2)/P3 (CH4)

Q= 0.01 x 0.0182/403= 5.06 x 10-11

 lnQ= -23.7                                                                                           

ΔG = 128.9 kJ/mol - 0.008314 kJ/mol K x 298K x 23.7  = 70.2 kJ mol

 

4. lnP = -kt + ln[P] = -3.0 x 10-6 x 3.0 x 105 + ln[2.4] = -0.0245

[P]= e-0.0245 = 0.98 atm

 

5. M NH4NO3= 28+4+48= 80 g/mol

4.25g/80 g/mol  = 0.053 mol

qrxn = -qsol

ΔH =qrxn  = -4.184 J/g oC x 64.25g (16.9 -22.0oC) = 1371 J (endothermic)

            ΔH= 1.371kJ/0.0531mol = 25.8 kJ/mol

 

  1. (a) ΔHorxn (kJ/mol)       = 2(-395) –2(-297) = -196 kJ/mol

ΔSorxn (J/mol)  = -191 J/mol K = -0.191 kJ/mol K

ΔGrxn (kJ/mol) = -196 kJ/mol -373K(-0.191 kJ/mol K) = -124.8 kJ/mol

lnKeq = - ΔGrxn /RT = 124.8 kJ/mol / 0.008314 kJ/mol K   x 373 K = 40.2

Keq = e40.2  =  2.87 x 1017 at 100 oC

(b)    T= 196/0.191 = 1026 K

(c)     Increase

 

  1. (a) exp 2 and 1: [C] = [B] = constant, 2 fold increase in  [A] no change  in the rate

zero order in A

exp. 2 and 3: [C] is constant;  2-fold increase in  [B] caused 2-fold increase in the rate: [2]m = 2; m=1; rxn is 1st order in  [B]

exp. 4 and 3 {B]=constant, 2 fold increase in [C] causes 4-fold increase in rate:

rxn is second order in C                 rate = k [C]2[B]

(b) overall rxn order:    2+1 = 3

(c) rate constant, k

0.0115 M/s = k (10-2 M)2 (10-2 M)= k 10-6 M3

k = 0.0115 Ms-1/ 10-6 M3 = 0.0115 x 106 = 1.15 x104   s-1M-2

d)   rate= 1.15 x104   s-1M-2 (0.1M)3 = 1.15 x104 x 10-3 = 11.5  M s-1