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Laabri

Year 13 Entropy

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Last updated over 1 year ago
22 Nsɛmmisa
Hyɛ no nsow a efi ɔkyerɛwfo no hɔ:

Edexcel chemistry entropy using system and surroundings

1
1
5
1
1
1
1
1
1
1
1
1
1
1
1
2
3
1
1
1
1
1
Asemmisa {{asɛmmisaAhyɛnsode}}
1.

Asemmisa {{asɛmmisaAhyɛnsode}}
2.

Hint MP = 801oC

Boiling point = 1413oC

Asemmisa {{asɛmmisaAhyɛnsode}}
3.

Explain why the entropy of the system increases when sodium chloride dissolves in water. (2)

In the solid / lattice, the ions are (1) But in the solution the ions are /more dispersed / more spread out/ more random (so entropy has increased) (1)

(Dissolving results in) an increase in the number of moles of particles and from to / NaCl → Na+ + Cl−

(1) The bonds between the water molecules are disrupted

Mmuae Afoforo a Wobɛpaw:
one
hydrogen
fixed
mobile
two
Asemmisa {{asɛmmisaAhyɛnsode}}
4.

(High) positive value is expected because:

moles → moles.

(two) → and/or liquid (+ one solid)

Entropy of the surroundings for an endothermic reaction is (1)

The reaction is feasible with the total entropy being (so entropy of the system has to be a large enough positive value for this to be true).

Mmuae Afoforo a Wobɛpaw:
3
negative
13
solids
positive
gas
Asemmisa {{asɛmmisaAhyɛnsode}}
5.

∆Sɵ system = (sign) JK-1mol-1

Asemmisa {{asɛmmisaAhyɛnsode}}
6.

sign

value = kJ mol-1

Asemmisa {{asɛmmisaAhyɛnsode}}
7.

sign=

value = JK-1mol-1

sigh =

value = J K -1 mol-1

Asemmisa {{asɛmmisaAhyɛnsode}}
8.

∑Sɵ (products) = +

∑Sɵ (reactants) = +

∆Sɵ system =

Asemmisa {{asɛmmisaAhyɛnsode}}
9.

(iii) Give two reasons why the positive sign of your answer to (a)(ii) is as you would expect.

more moles of than (3 moles → 13 moles)

Two solids in the → a gas / a liquid (+ 1 solid) in the

Mmuae Afoforo a Wobɛpaw:
products
reactants
Asemmisa {{asɛmmisaAhyɛnsode}}
10.

sign

value = J mol-1 K-1

Asemmisa {{asɛmmisaAhyɛnsode}}
11.

∆Sɵ becomes less negative (smaller in MAGNITUDE (because you are dividing −∆H by a larger T)

∆Sɵ / ∆H are not (significantly) affected by a change in temperature.

∆Sɵ increases

Mmuae Afoforo a Wobɛpaw:

total

system

surroundings

Asemmisa {{asɛmmisaAhyɛnsode}}
12.

(give your value as a decimal to 4 s.f.)

Asemmisa {{asɛmmisaAhyɛnsode}}
13.

Barium hydroxide will (very) soluble / will be sparingly soluble

and K value suggests that the equilibrium lies to the side / reactants

OR K << 1 so predominant

Mmuae Afoforo a Wobɛpaw:
products
be
right-hand
left-hand
not be
reactants
Asemmisa {{asɛmmisaAhyɛnsode}}
14.

M1: Hydroxides get soluble as you descend Group 2 (1)

M2: ∆Sɵ total gets negative as you go from Ca(OH)2 to Ba(OH)2

Mmuae Afoforo a Wobɛpaw:
more
more
less
Asemmisa {{asɛmmisaAhyɛnsode}}
15.

J mol−1 K−1 Note the units are J not kJ

Asemmisa {{asɛmmisaAhyɛnsode}}
16.

and solid react to to form

There are fewer ways of arranging particles in a than a .

mol(es) of reactant forming mole of product.

mol(es) of products than reactants

Mmuae Afoforo a Wobɛpaw:
solid
one
Two
Fewer
gas
Asemmisa {{asɛmmisaAhyɛnsode}}
17.

Give the formula to be used

= +

Mmuae Afoforo a Wobɛpaw:

Asemmisa {{asɛmmisaAhyɛnsode}}
18.

Sign =

value = J mol−1 K-1

Asemmisa {{asɛmmisaAhyɛnsode}}
19.

value = - kJmol−1 Note the units

Asemmisa {{asɛmmisaAhyɛnsode}}
20.

Asemmisa {{asɛmmisaAhyɛnsode}}
21.

sign

value kJmol−1

sign

value J mol−1 K−1 Note the units

sign =

value = J mol−1 K−1

Asemmisa {{asɛmmisaAhyɛnsode}}
22.

Decrease in T: is not (significantly) changed.

or (−∆H/T) is more positive.

makes more positive

Mmuae Afoforo a Wobɛpaw:
∆Stot

∆Ssystem

∆Ssurr