Could anyone help me with couple of these questions?

Both the questions have been uploaded to this post.

Please provide a brief explanation if possible with the answer. Thanks

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What have you tried?

I haven't done A level chemistry for over 5 years. From memory and some reasonable knowledge of the equilibrium constants, i worked out the answer for question 8 as Exothermic and the temperature increasing. Not sure if i am entirely correct?

I do not have an idea how to start about answering question 9.

Do you have an idea?

(edited 7 years ago)

Reply 3

Saint132

I think i have figured out the Ka value and i have got it as B (5 ka). Am i correct?

Reply 4

Pigster

Kc increases with temperature. Kc = [products]/[reactants]. Increasing Kc means more products than before, i.e. equilibrium -> RHS. Therefore endothermic. Q9: I'd go with Ka = [H+] x [A-]/[HA]. When you add half the volume of OH- needed to fully neutralise (i.e. the volume that makes the graph go vertical) then [A-] = [HA], so Ka = [H+] and pKa = pH = 5.

Reply 5

NeverLucky

Original post by Saint132

I think i have figured out the Ka value and i have got it as B (5 ka). Am i correct?

Yes,

pK_a=pH

at the half-equivalence point. Since the equivalence point is

30 cm^3

, the half-equivalence point is

15 cm^3

which corresponds to

pH=5

pK_a=5

. Thus

Unparseable latex formula:

K_a=10^-^5

Reply 6

Saint132

Yes. Thank you for confirming!

Question 8, i think i have made a mistake and the answer should be in fact be D (Increases and it should be endothermic as Kc has increased due to the temperature being increased)?

That is from what i have read. Is this correct also?

Reply 7

NeverLucky

For question 8:

K_c = \frac{[NO]^2} {[O_2][N_2]}

So as the value of

K_c

has increased when the temperature has increased, it means the

[NO]

has increased in relation to the

[N_2]

and

[O_2]

. Therefore, the forward reaction has to be endothermic so equilibrium would shift to the right to oppose the increase the temperature thus increasing the concentration of

NO

Reply 8

Saint132

One final question. I have worked out question 10 but question 11, i have no recollection of the theory.

Could you help with question 11 if possible.

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Reply 9

Saint132

Original post by NeverLucky

For question 8:

K_c = \frac{[NO]^2} {[O_2][N_2]}

So as the value of

K_c

has increased when the temperature has increased, it means the

[NO]

has increased in relation to the

[N_2]

and

[O_2]

. Therefore, the forward reaction has to be endothermic so equilibrium would shift to the right to oppose the increase the temperature thus increasing the concentration of

NO

Yeah. Thanks again. I think i should have done a chemistry degree rather than a physics related one as i had more intrigue in the last day working these out than any of my physics topics.

Reply 10

NeverLucky

Original post by Saint132

One final question. I have worked out question 10 but question 11, i have no recollection of the theory.

Could you help with question 11 if possible.