synthesisFR
Well-Known Member
i got D but answer says C?
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Chem (2 Viewers)
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i got D but answer says C?
synthesisFR
Well-Known Member
would u say that change in entropy > 0? because like the moles is greater in products but theres less gas, and gas molecules r usually much more entropic than water?
yeah you’re right
synthesisFR
Well-Known Member
is this the best way to do part c ^?
i did MM whole polymer - (2n-1)H20 where n is 1000
why did i get teh wrong answer with that
synthesisFR
Well-Known Member
can someone give me a sample answer for Baulko chem 2020 q23 part E?
their answer doesn't rlly make sense and im not sure how to describe the different values for the concentration. like ur using two different methods but idk how to explain the diff between results
their answer doesn't rlly make sense and im not sure how to describe the different values for the concentration. like ur using two different methods but idk how to explain the diff between results
synthesisFR
Well-Known Member
help
synthesisFR
Well-Known Member
for this we do Q = -nxenthalpy change
but we just find the one which produces the largest moles of water?
but what if the enthalpy change is larger for one of the reactions ?
synthesisFR
Well-Known Member
shouldn't part a be 3dp bc u actually get 3.523 and lowest sf is 3
shouldnt part b be 3sf
@carrotsss
synthesisFR
Well-Known Member
shouldnt it be 2.40 x 10^-5
bc u get 2.3979... x 10^-5
synthesisFR
Well-Known Member
shouldnt it be both 1 sf bc the ph is to 1dp @carrotsss
no trial paper follows these rules and each trial paper makes up their own mf rules when it comes to ph i wanna die
synthesisFR
Well-Known Member
when we draw structural formula tho we put a negative on top of O and positive on top of Na but here we dont need to?
This question comes from the 2021 NEAP paper. Some versions floating around have a note added that the correct answer is D rather than C.
Both methods should work, and give the same result:
n HOOC-(CH2)3-COOH + n H2N-(CH2)2-NH2 ---> HO-(-OC-(CH2)3-CO-NH-(CH2)2-NH-)n-H + (2n - 1)H2)O
Thus, the polymer is HO-(-OC-(CH2)3-CO-NH-(CH2)2-NH-)n-H = HO-(C5H6O2-C2H6N2)n-H, and so:
By the method used in the solutions:
M(polymer) = 17.008 + (98.098 + 58.088) x n + 1.008 = 18.016 + 156.186n g mol-1
By the conservation of mass method:
M(polymer) = n x (132.114 + 60.104) - (2n - 1) x 18.016 = 192.218n - 36.032n + 18.016 = 156.186n + 18.016 g mol-1
Good point. The first three answers are strong acid / strong base and so the reaction is the same - hydronium plus hydroxide - and has the same enthalpy change. The reaction in (D) involves a weak acid and so the reaction will also occur as ethanoic acid plus hydroxide to make ethanoate ions plus water. This has a smaller magnitude enthalpy change as an O-H covalent bond is broken, which is endothermic.
synthesisFR
Well-Known Member
help with this needed and all the other posts i made below this one ^
thx
i don’t think the answer is right because like you can account for the non dissociated component with Ka
It's not just using two methods, but measuring two different (though related) values.
Part (d) is calculating the concentration of unionised ethanoic acid molecules in the solution at equilibrium.
The titration calculation in part (a) is calculating the concentration of unionised ethanoic acid molecules as if the system on the LHS of the equilibrium were just mixed and before the forward reaction first occurs. This is because the titration involves reactions of the hydroxide ions with both the unionised ethanoic acid molecules and with the hydronium ions produced by the ionisation. Alternatively, you can view this as the presence of hydroxide ions driving the ionisation process to completion by removing the hydronium product as it forms.
In theory, answer in part (a) = [CH3COOH]initial = [CH3COOH]eqm + [CH3CO2-]eqm = answer in part (d)
which is true to the limit of accuracy available, as both are 1.4 M (2 sig. fig.).
I think it’s relatively similar for all of these reactions
(A) n(H+) = 0.0050 mol and n(OH-) = 0.080 mol ===> H+ is limiting ===> delta_q = -0.0050 x -56 = +0.28 kJ
(B) n(H+) = 0.010 mol and n(OH-) = 0.10 mol ===> H+ is limiting ===> delta_q = -0.010 x -56 = +0.56 kJ
(C) n(H+) = 0.020 mol and n(OH-) = 0.080 mol ===> H+ is limiting ===> delta_q = -0.020 x -56 = +1.12 kJ
(D) IF it was strong / strong: n(H+) = 0.050 mol and n(OH-) = 0.040 mol ===> OH- is limiting ===> delta_q = -0.040 x -56 = +2.24 kJ
These calculations assume an enthalpy of neutralisation of -56 kJ mol-1, but the same could be found taking this as an unknown value X.
The answer is (D) provided that the enthalpy of neutralisation of ethanoic acid with hydroxide is below -28 kJ mol-1, which is likely as some of the reaction is still hydronium and hydroxide and the reaction with hydroxide and unionised ethanoic acid molecule do break an O-H bond (endothermic) but also replace the ion-dipole interactions of hydroxide and water with new ones for ethanoate and water, which partially compensates for the O-H bond breaking. Enthalpy of neutralisation for a weak acid with a strong base will be lower in magnitude for weaker acids and when the covalent bond to H being broken is stronger.
Yes, in (a) you have 3 sig. fig. for all data, so you can assume 3 dec. pl. in the pH.
In (b), I would give 3 sig. fig., but technically the pH = 7.0 only entitles you to 1 sig. fig. It would be clearer if part (b) said to bring the mixture "back to neutral", which would also avoid a need to specify a temperature.