[777888]

7. For the rxn: A + B + 200kJ <-> C + D, predict the effects after drecreasing [D].
a)release energy
b)increase [C]
c)both a and b

[b is definitely right, but I'm not quite sure about a...the temperature would decrease, but the 200kJ is in the left side of the rxn...]

[Donaldson Tan]

haha.. the tabulation method i taught u is now called ICE. Cute name..
Normally, exam-wise, if you arent given the Ksp of Ca(OH)2, you can safely assume 100% dissociation.

what is dH for question(7) ?

[777888]

haha.. the tabulation method i taught u is now called ICE. Cute name..
Normally, exam-wise, if you arent given the Ksp of Ca(OH)2, you can safely assume 100% dissociation.

what is dH for question(7) ?

what is dH

[Donaldson Tan]

7. For the rxn: A + B + 200kJ <-> C + D, predict the effects after drecreasing [D].
a)release energy
b)increase [C]
c)both a and b

what is deltaH for the reaction in question 7?

[777888]

what is deltaH for the reaction in question 7?

+200kJ, but according to Le Chatelier's principle, the 200kJ will decrease...

[Donaldson Tan]

deltaH of a reaction is different from applying Q of 200kJ to the reaction system.

[777888]

deltaH of a reaction is different from applying Q of 200kJ to the reaction system.

But after the stress and the equilibrium shift, the temperature of the SYSTEM will decrease(because heat energy decreased), right? (releasing energy?!)

[777888]

Let me make an example that is similar.
5.20mL of 0.20M HCN is titrated with 0.35M NaOH. Find
a)pH initially and
b)at midpoint(halfway to equivalence).

I calculated this way:
a)HCN+NaOH->H2O+NaCN
HCN+H2O<->H3O+ + CN- (hydrolysis)
Ka=6.2x10^-10=x^2 / 0.2 (assumption made)
pH=4.95

b)HCN +   NaOH->H2O+NaCN
n=0.004    n=0.002
V=0.02      c=0.35
V of NaOH required to reach midpoint=0.002/0.35=0.00571

After rxn:
HCN +   NaOH->H2O+NaCN
n=0.002  n=0            n=0.002
c=n/V(total)
c=0.1556                   c=n/V0.07779

HCN+H2O<->H3O+ +CN-(hydrolysis)
6.2x10^-10=0.07779x/0.1556 (assumption made)
pH=9.21
The results are surprising to me...the pH is so high at midpoint...all other weak acid/strong base quesitons I did before have midpoint at about 5 point something only. So did I calculated them right?

[777888]

8.Which of the following is in a state of equilibrium
a)A container with 20mL of water
b)A container with an ice cube and water
c)both a and b
[would a container with 20mL of water be in a state of equilibrium]

[Donaldson Tan]

(7) I'm not sure of your notation. I assume "..+200kJ" means u add 200kJ of heat, and it's not deltaH of the reaction. Adding heat to the system favours the endothermic reaction.

( for (a). H2O(l) <-> H2O (g), but for (b), there will be heat transfer from water to the ice that cause it to melt, hence no equilibrium.

[777888]

(7) I'm not sure of your notation. I assume "..+200kJ" means u add 200kJ of heat, and it's not deltaH of the reaction. Adding heat to the system favours the endothermic reaction.

( for (a). H2O(l) <-> H2O (g), but for (b), there will be heat transfer from water to the ice that cause it to melt, hence no equilibrium.

7)Yes the reaction is endothermic:
A + B + heat <-> C + D
and then [D] is decreased, so shift right and [A], and heat all decreased. But heat decreased means temperature is lowed and is releasing energy, right? (heat energy is consumed...)

8b)H2O(l) <-> H2O(s) why not equilibrium? I thought that there will be equilibrium of melting and solidfying processes occuring at the same rate...

[Donaldson Tan]

( Common sense lah, u put ice in water, what do u see? however, u find liquid water exist without any significant change in its mass..

[777888]

Let me make an example that is similar.
5.20mL of 0.20M HCN is titrated with 0.35M NaOH. Find
a)pH initially and
b)at midpoint(halfway to equivalence).

I calculated this way:
a)HCN+NaOH->H2O+NaCN
HCN+H2O<->H3O+ + CN- (hydrolysis)
Ka=6.2x10^-10=x^2 / 0.2 (assumption made)
pH=4.95

b)HCN +   NaOH->H2O+NaCN
n=0.004    n=0.002
V=0.02      c=0.35
V of NaOH required to reach midpoint=0.002/0.35=0.00571

After rxn:
HCN +   NaOH->H2O+NaCN
n=0.002  n=0            n=0.002
c=n/V(total)
c=0.1556                   c=n/V0.07779

HCN+H2O<->H3O+ +CN-(hydrolysis)
6.2x10^-10=0.07779x/0.1556 (assumption made)
pH=9.21
The results are surprising to me...the pH is so high at midpoint...all other weak acid/strong base quesitons I did before have midpoint at about 5 point something only. So did I calculated them right?

OK! Thanks!
How about the titration calculations? Did I do it right?(espically the hydrolysis part...)