Exam 1 Practice
--------------------------------------------------------------------------------------------------------

Log(k1/k2) = Ea(1/T2-1/T1)/(2.303R) k = AeEa/RT R = 8.314joules/(Kxmol)

log[A]t = -kt/2.303 + log[A]o ln[A]t = -kt + ln[A]o t1/2 = .693/k

rate = D[A]/Dt = -k[A]x

--------------------------------------------------------------------------------------------------------

Ka(acetic acid)= 1.8x10-5 Kb(NH3)=1.8x10-5

Ka1Ka2=KaH2S=9.1x10-21 Kw=[H+][OH-] = 1.00x10-14

pH=-log[H+] pOH=-log[OH-] pH + pOH =14 Kw=KaKb

-----------------------------------------------------------------------------------------------------------
1        The hydrolysis reaction,
(CH3)3CCl + H2O ---> (CH3)3COH + H+ + Cl-

is carried out at two temperatures and the following data are obtained.

20° C 45° C .

Exp.    [(CH3)3CCl]    [H2O]      Rate                    [(CH3)CCl]       [H2O]      Rate

1          .0100               20.0        2.41                       .0100              20.0        6.85
2          .0300               20.0        6.41                       .0300              20.0        20.5
3          .0500               20.0        12.1                       .0500              20.0        34.3
4          .0100               40.0        2.40                       .0100              40.0        6.86
a) Write the rate expression for this reaction.

 
Rate=k[(CH3)3CCl]1[H2O]0
 

 

b) Calculate the rate constants at both temperatures for the reaction.

 
k20º= rate/[(CH3)3CCl]=2.41/.01=241

k40º= rate/[(CH3)3CCl]=6.85/.01=685
 

c) Calculate the activation energy for the reaction.

 
Ea=(logk20º/k40º)2.303R/[1/T40º-1/T20º]=(log241/685)2.303x8.314J/Kmol/[1/318K-1/293K]
 

 
 
 

2        What is the slope of the concentration [(CH3)3CCl] vs time from the first experiment at 20°C in problem number 1? __-2.41________
(Hint: No calculations are needed. Careful of the sign.)

 
 

3        Consider the mechanism:
2NO <===> N2O2 Fast
N2O2 + H2 ---> N2O + H2O Slow

N2O + H2 ---> N2 + H2O Fast

a) What is the overall reaction?

 
2NO + 2H2 -->N2 + 2H2O

 

b) What are the intermediates?

N2O2
N2O

 

 

c) What is the rate expression for the reaction?

  rate=k[N2O2][H2] for slow step
Keq=[N2O2]/[NO]2  ==> [N2O2]= Keqx[NO]2
rate=kKeqx[NO]2[H2]
  rate=k[NO]2[H2]
 

4       Consider the equilibrium system: CH4(g) + 2H2S(g) <==> CS2(g) + 4H2(g) + heat
a) At 900°C Kc= 3.59. What direction will the reaction shift if [CH4]=1.50 M,
[H2S]=2.00 M, [CS2]=.500 M, and [H2]= 3.00 M.
 
 
Q=[CS2][H2]4/[H2S]2[CH4]=(.5)(3)4/(2)2(1.5)=6.75  since Q>Kc the reaction will shift left.
 

 
b) If the reaction is at equilibrium, in what direction would reaction shift if you:

a removing CH4?           d. increasing the pressure on the system?

______Left________       _______Left_________

b. adding H2S?              e. adding Ni catalyst?

______Right_______       ______No Shift_________

c. cooling?                      f. removing CS2(g)

________Right_____      _____Right____________
 

5         An empty 1.00 liter container is filled with 2.00x10-2 mol of H2 and 1.00x10-2 mol of I2 at 220°C and allowed to come to equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 4.56x10-3M. Calculate Kc at 220°C for the reaction:
H2(g) + I2(g) ---> 2HI(g)

 

  Kc=(.00456)2/[(.02-.00456/2)(.01-.00456/2)]
 

6        Sketch the graph of concentration vs time for the reactants and products of the reaction as described in problem 5. H2(g) + I2(g) ---> 2HI(g)
|

|

|

|

|

[ ]             |
|

|

|

|

|________________________________________|________________________

0                      >--- Time --->                         tequilibrium
 

7      What is the pH of a .500 M solution of ammonium chloride? Kb=1.80x10-5  
pH=4.8
 

 

8         Calculate the pH of a solution of a .500 M solution of ammonia? Kb=1.80x10-5
  pH=11.5
 

9        20.0 mL of .100 M NH3 is added to 10.0 mL of .100 M HCl. What is the pH of the resulting solution? NH3(aq) + HCl(aq) ---> NH4+(aq) + Cl-(aq)
  pH=9.25
 
10        Design a buffer having a pH of 10. Give the concentrations of each component.

 
    [NH3]=1M  and [NH4+]=.18 would work swimmingly.
 

 
 
 

11       What is the pH of a 2.00 M solution of sodium acetate?
 

 
pH=9.52
 

 

12        Determine whether the following substances will be acidic, basic or neutral in water solution.
a. KCN Acidic {Basic} Neutral           b. AlCl3 {Acidic} Basic Neutral

c. KOAc Acidic {Basic} Neutral         d. NH4Br {Acidic} Basic Neutral

e. NH4CN Acidic {Basic} Neutral       f. NH4OAc Acidic Basic {Neutral}

g. NaClO4 Acidic Basic {Neutral}       h. KH2PO4 {Acidic} Basic Neutral

i. CH3OH Acidic Basic {Neutral}       j. KNO3 Acidic Basic {Neutral}
 

13       Calculate the [PO4-3] in a 3.00 M phosphoric acid sloution.
 
K1K2K3=Ka=[H+]3[PO4]/[H3PO4]
Remember that all of the [H+] comes from the first dissociaiton. If you write the equilibrium expression for the first dissociation, and solve for the [H+] you get:
[H+]={K1[H3PO4]}.5
Solving the first expression for phosphate and plugging in the above [H+] yields:
[PO4]=Ka[H3PO4]/ [H+]3=Ka[H3PO4]/ ({K1[H3PO4]}.5)3

  Ka(3.00)/ ({K1(3.00)}.5)3 You'll need to look up K1 

 

14       Determine whether the following acids and bases are strong or weak; Ka>1 or Ka<1.
a. HClO Strong {Weak}           b. HClO4 {Strong} Weak

c. H2SO4 {Strong} Weak        d. K2S Strong {Weak}

e. HCN Strong {Weak}           f. HI {Strong} Weak

g. HNO2 Strong {Weak}         h. KF Strong {Weak}

i. HF Strong {Weak}               j. H3PO4 Strong {Weak}

15      Show the Ka expressions for HCN and NH4+.

     Ka=[H+][CN-]/[HCN]         Ka=[H+][NH3]/[NH4+]
16       Show the Kb expressions for CN- and NH3.
     Kb=[HCN][OH-]/[CN-]         Ka=[NH4+][OH-]/[NH3]