1.         2H2O(l)   +  heat  ⇌    H3O+   +         OH-                 Kw       =1.0  x  10-14

 

The value of the Kw decreases as temperature decreases because the endothermic reaction will shift to the left as the temperature decreases.

 

 

2.         a)         [HI] was increased and the equilibrium shifted left.

 

            b)         The volume was increased and all concentrations increased with no                                 shift.

 

            c)         The temperature was increased and the equilibrium shifted to the left.

 

            d)         [I2] was decreased and the equilibrium shifted to the left.

 

3.         2NO(g)             +          O2(g)                 ⇌        2NO2(g)

 

           

The rate of the forward reaction starts high and gradually decreases until it is equal to the reverse rate. As reactants convert into products, the concentration of the reactants decreases and there are less reactant collisions.

 

 

4.         CaSO3(s)          ⇌        CaO(s)  +          CO2(g)

 

If the volume decreases to half the initial volume, the concentration of CO2 would double and then decrease as the equilibrium shifts to the left.

 

 

 

 

[CO2]
 
 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


5.                     CO(g)    +          H2O(g)             ⇌        CO2(g)  +          H2(g)                                       Keq     =          16

 

 

            I           0.30 M            0.30 M                        0.90 M            0.90 M                        Trial Keq        =            (0.90)2             =          9

                                                                                                                                                                                    (0.30)2

            C         - x                    - x                                + x                   + x                                                       Kt                    Keq

                                                                                                                                                                                          

            E          0.30 - x            .30 - x                          0.90 + x           0.90 + x                                               9                      16

 

                                                                                                                                                Trial Keq  < Keq      Shifts Right

 

            Keq     =          (0.90  +  x)2     =          16

                                    (0.30  -  x)2

 

 

                                    0.90  +  x         =          4

                                    0.30  -  x

 

 

                                    0.90  +  x         =          4(0.30  -  x)

 

                                    0.9  +  x           =          1.2  -  4x

 

                                    5x        =          .3

 

                                    x          =          0.060 M

 

                                    [CO2]   =          0.90 M  +  x

 

                                    [CO2]   =          0.90 M +  0.06 M

 

                                    [CO2]   =          0.96 M

 

 

6.         CO(g)                +          2H2(g)                ⇌        CH3OH(g)        +          energy

 

           

 

 

a)         When the volume decreases all concentrations increase so [CH3OH] increases.

           

            When the volume decreases the pressure increases and the equilibrium shifts to the side with fewer gas particles, which is the products, so    [CH3OH] increases.

 

b)         Nothing happens to [CO] because a catalyst does not shift an equilibrium.

 

7.                                 SnO2(s) +          2H2(g)                ⇌        Sn(s)     +          2H2O(g)            Keq     =          1.00

 

 

                        I                                   0.300 M                                              0                      Leave out the solids and divide by 3.0 L to get M.

                       

                        C                                 x                                                          x

 

                        E                                  0.300  -  x                                            x

 

           

                        Keq     =          [H2O]2

                                                           

                                               

                                                [H2]2   

 

 

                        Keq     =          x2

                                                                        =          1.00

 

                                                (.3  -  x)2

 

                                                x

                                                                        =          1.00

 

                                                (.3  -  x)

 

                                                x          =          .3  -  x

 

                                                2x        =          .3

 

                                                x          =          0.150 M

 

                                                [H2]     =          0.300 M  -  0.150 M

 

                                                            =          0.150 M                                  Not too shabby!

                                   

 

                                                3.00 L H2        x          0.150 moles    x          1 mole Sn        x          118.7 g            =            26.7 g     We are good!

                                                                                    1 L                               2 moles H2                  mole

 

 

8.                                 Fe3+(aq)             +                      SCN-(aq)                       ⇌                    FeSCN2+(aq)

                                                                                                                                                                          Note the two solutions dilute each other

                        25        0.600 M                      25        0.400 M

                        50                                            50                                                                                            Note the loss of 1 significant digit, hello............

 


            I                       0.300 M                                  0.200 M                                              0

 

            C                     0.190 M                                  0.190 M                                              0.190 M

 

            E                      0.110 M                                  0.010 M                                              0.190 M

                       

 

                                    [FeSCN2+]

            Keq     =                                 

                       

                                    [Fe3+][SCN-]

 

            Keq     =          (0.190)                         =          1.7  x  102                    Awesome work!!

                                    (0.110)(0.010)

 

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               

 

 

9.         In experiment 2 compared to experiment 1 the temperature is higher and the [CO] has increased and the [CH3OH] has decreased. This           means that the equilibrium has shifted towards the reactants. An exothermic reaction would shift to the reactants when the temperature was    increased. I disagree and the student must be dumb!

 

10.       System One                                                                System Two

            HInd(aq)           ⇌        H+(aq)   +          Ind-(aq)              2CrO42-(aq)       +          2H+(aq)             ⇌        Cr2O72-(aq)            +          2H2O(l)

            yellow                                                  orange             yellow                                                              orange

 

            Adding HCl increases the [H+], which would shift the first reaction left and the second reaction right. The unknown equilibrium must be            system Two as it would change from yellow to orange as it shifted to the right.

 

11.       The key to this one is to make an ICE chart.

 

                        N2O4(g)             ⇌        2O2(g)                +          N2(g)

           

            I           2.00 M                        0                                  0

 

            C         0.40 M                        0.80 M                        0.040 M          Do not forget to divide by 2.................Hello..............

 

            E          1.60 M                        0.80 M                        0.40 M            One Significant figure is lost in the ICE chart                2.00 M

                                                                                                                                                                                                            -           1.60 M

                                                                                                                                                                                                            =          0.40 M

 

                                    [O2]2[N2]                                             (0.80)2(0.40)

            Keq     =                                              =                                                          =          0.16

                                   

                                    [N2O4]                                                 (1.60)

 

 

12.                   CO(g)                +          H2O(g)              ⇌        CO2(g)              +          H2(g)

 

                       

            E          0.20 M                        0.60 M                        0.60 M                        0.60 M

 

                       

                                                            [CO2][H2]

                                    Keq     =                                              =          (0.60)(0.60)     =          3.0

                                                                                                            (0.20)(0.60)

                                                            [CO][H2O]

 

 

           

After 2.00 moles CO is added to the initial 1.00 moles there are 3.00 moles divided by 5.00 L or 0.60 M and the reaction will shift to the right to get to equilibrium.

                       

 

 

                        CO(g)                +          H2O(g)              ⇌        CO2(g)              +          H2(g)

 

                       

            I           0.60 M                        0.60 M                        0.60 M                        0.60 M

 

            C         - x                                - x                                + x                               + x

 

            E          0.60  -  x                      0.60  -  x                      0.60  +  x                     0.60  +  x

 

 

                                                            [CO2][H2]

                                    Keq     =                                              =          (0.60  +  x)2     =          3.0

                                                                                                            (0.60  -   x)2

                                                            [CO][H2O]

           

 

                                                                                                            0.60  +  x         =          1.73205

                                                                                                            0.60  -   x

 

                                                                                                            0.60  +  x         =          1.73205(0.60  -  x)

 

                                                                                                            0.60  +  x         =          1.03923  -  1.73205x

 

                                                                                                            2.73208x         =          0.43923

 

                                                                                                            x                      =          0.16 M

 

                                                                                                [CO2]               =          0.60 M +  0.16 M       =          0.76 M        Yeaaaaaaaaaah!!!