Chem 127 exam IV     

November 17, 2004

           

Exam Questions-- With Answers

                       

Short Answers

1. (6 pts) define a catalyst (there are two important features of a catalyst)

           

A substance that (a) increases the rate of a chemical reaction and (b) is not conumed in the reaction-- that is, it is not a reactant.

 

2. (6 pts) Refer to the figure on the right. (Plots of ln k vs 1/T for several reactions)

            Curve A is the curve for the decomposition      

                        of hydrogen peroxide

            One of the other curves represents the

            decomposition of H2O2 in the presence of a

            good catalyst.

            Which curve is it, and how do you know that?

 

Since a catalyst sppeds up a reaction (kigher k at any temperature) it must be one of the upper curves (d-e-f)

A catalyst lowers the energy of activation for a reaction; it therfore must have a lower slope than curve A.

Thus it must be curve "f"

 

3. (6 pts) The thermodynamics table has 2 entires for Iodine (I2)

                                    DHof     DGof         So

                                    kJ/mol   kJ/mol   J/K mol

            I2 (s)                0            0            116.1

            I2 (g)                62.4       19.4       260.6

           

Therefore,  DGo = +19.4 kJ per mole for the reaction                I2(s) ---> I2(g)

            is the reaction spontaneous?   _NO, DGo>0______

                       

If you look in a bottle of I2 solid, you can see the purple vapor of I2(g) and you can smell the vapor.  Also, the solid vanishes all by itself if you leave the bottle open off for a day or two.

Assuming you said no (not spontaneous) how do explain the (spontaneous) presence of vapor and (spontaneous) sublimation that occurs?

 

This hinges on the definition of spontaneous and that little superscript 0. If all components are in the standard state, the reaction is not spontaneous. That is, if I2(gas) were at 1 atm, the reaction would proceed in reverse until some lower vapor pressure were reached. It does not mean that no vapor exists, just that the equilibrium is below the pressure of the standard state.

 

 

 

 

4. (6 pts) Consider the reaction   2C + O2 (g)  ----> 2 CO (g) 

          DH reaction = -221 kJ

            a. (5 points)  Do you expect DS for the reaction to be >0, <0, nearly 0 or is there no basis for a good guess? (WHY?)

Gases have higher entropy (disorder) than solids. Reactants are a solid (low entropy) and 1 mole of diatomic gas. Product is two moles of gas. This is a clear increase in entropy so DS >0.

 

5.(8 pts) Write the chemical equation and the equilbrium expression for

 

            a.  NH3 acting as a base

                        reaction is :

                        NH3 + H2O --> OH- + NH4+

 

                        Kb = [OH- ]* [NH4+ ] / [NH3]

 

            b. NH4+ acting as the conjugate acid (conjugate to the base NH3)

                        reaction is

                        NH4++ H2O -->H3O+ + NH3

 

                        Ka = [ H3O+ ] *[ NH3 ] / [NH4+]

 

            c. The solubility of the salt  Ag2SO4

                        reaction is:   Ag2SO4 (s) --> 2 Ag+ + SO42-

 

                        Ksp = [Ag+ ]2 * [ SO42- ]

 

            d. What is the property of a pH buffer that makes it useful (makes it a buffer?)

 

A buffer resists change in pH-- on dilution and on the addition of modest amounts of acid or base. (That is, the pH changes by realtively small amounts under these conditions.)

 

 

6.5 (2 points) Why is it impolite to look a gift horse in the mouth ? (Monday’s class)

            a. It embarrases the horse who is likely to be shy

            b. It suggest you are second guessing the age of the horse (you'd know

                        this if you attended class on Monday.)

            c. You should be keeping your eye on the person making the gift

                        since you probably can’t trust them

            d. horses are marked on the teeth so you are checking to see if this horse has been stolen          

 

Problems

7. (6 points) No work needs to be shown for this problem-- answers only

            What is pH of a 0.025M Ca(OH)2 solution?

Ca(OH)2 is a strong base so the [OH-] = 0.050 M

pOH = - log10(0.050) = 1.30

pH = 14.0 - 1.30 = 12.70

 

 

            What is [H3O+] for a pH 6.86 Buffer?

[H3O+] = 10-6.86 = 1.38 x10-7

 

 

8.a (20 pts) What is the pH at the equivalence point in the titration of 0.10 M NaOH

            with 0.10 M Acetic Acid?         (Ka of Acetic Acid is 1.80 x 10-5.)

 

At the equivalence point all of the NaOH has reacted (with the added Acetic Acid) annd none of either remains. The prodcut is simply a solution of acetate ion. The concentration is 0.050 M (because the volume has doubled due to added titrant.)

 

Acetate is a weak base, the conjugate base of acetic acid

            Kb = 1.0 x 10-14 / Ka = 5.56x10-10

            [OH=] = (apprx) = sqrt (Kb * C) = (5.56 x 10-10 * 0.050)1/2 = 5.27 x 10-6

            pOH= -log( that value) = 5.72

            pH = 14.00 - 5.72 = 8.72

 

 

 

8b. Sketch the pH vs volume    for this titration curve.. Make reasonable estimates of pH

            We start with 0.1M NaOH, a strong base (pH= 13)

            As we add acetic acid, the pH changes slowly since the pH is fixed by the stronger base (OH-). For example, when half of the OH- disappears, the pH decreases by log(2) or 0.30.

            As we near the equivalence point the pH drops sharply (to the 8.7 we computed above)

            Beyond the equivalance point we have a mixture of Acetic acid and socium acetate so the pH remains fairly close to the pHa of acetic acid (4.7)

 

9. (20 pts) What is Kp at 25oC for the reaction

          H2 (g) + I2 (g) ----> 2 HI (g)

                                    DHof     DGof         So

                                    kJ/mol   kJ/mol   J/K mol

            H2 (g)              0            0            130.6

            I2 (g)                62.4       19.4       260.6

            HI(g)               26.5     1.7           206.5

First, evaluate DG0 for the reaction

            DGo = 2( 1.7) - (0) - (19.4) = -16.0 kJ = -16 000 J

            DG = - RT ln (Kp)

            DGo/(-RT) =  + 16000/ (8.314*298) =+ 6.45

            Kp = (anti ln) (6.45) = exp (6.45) =637.7

           

 

 

 

 

 

 

 

10. (20 pts) The decomposition reaction of Hydrogen Peroxide, H2O2, is studied in the presence of H+ and permanganate ion..

                   2 H2O2 (aq) --- 2 H2O  + O2 (g)

 

 *. rate is moles/liter-sec  (change in the concentration of H2O2.)

            a. What is the order of ther eaction with respect to H2O2, H+ and MnO4- ?

            b. What is the rate constant for the reaction ?

 

run #

Initial   rate (*)

Initial [H2O2]

initial [MnO4-]

initial  [H+]

1

0.0045

0.10 M

.00010 M

.018M

2

0.0023

0.050 M

.00010 M

.018 M

3

0.0013

0.050 M

0.00005

.018 M

4

0.0005

0.10 M

0.0001

0.006 M

 

First, the table had 3 errors, corrected in bold type. The last error was found too late to correct, so we will grade the logic of what you found. (You may actually find a negative order for H+ due to the error) The discussion below interprets the intended data.

            The reaction rate is affected by MnO4- and H+ but they are not reactants (catalysts)

            Comparing runs 1 and 2-- all is the same except the [H2O2]

                        decreasing [H2O2] by a factor of 2 decreased the rate by 0.0045/.0023= 1.95

                        that's the prediction for a first order reaction

                        thus, the reaction is first order in peroxide

            Compare runs 2 and 3-- all is the same except the MnO4-

                        decreasing permanganate by 2 fold decreased the rate by .0023/0.0013 = 1.77

                        that's close to the predicted 2 fold for first order

                        thus reaction is also first order in permanganate

            Now compare run 4 with run 1-- only H+ cahnges

                        the concentration decreases by a factor of 3

                        the rate decreases by a factor of 9  (that's 32)

                        so reaction is second order in H+

 

Now rate = k [H2O2]1  [MnO4-]1 [H+]2

Use any row of data, say row 1

            .0045 = k * 0.010 * 0.0010 * (0.018)2

            k = 1.39 x 10+6

 

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