1 a +788 kJ mol-1
b This is the reverse or the equation for lattice enthalpy formation, so the sign of 6.H is changed.
c The enthalpy change of lattice dissociation.
2 a The electron is auracted by the sodium nucleus, so energy must be put in to remove it.
b The electron is attracted by the chlorine nucleus so energy is given out during this process.
3 a i Al(g) → Al+(g) + e–
ii
AJ+(g) -+ Al2+(g) + e–
1 a
b @
-929 kJ mol-1
1
+22 kJ mol-1
Small because the energy put in to break the lattice is of similar size to that given out when the ions arc hydrated.
2
They are small and highly charged positive ions so they strongly polarise negative ions. The calculated value would be greater because there is extra covalent bonding.
1 a i Mg(s) + ZnO(s) -» MgO(s) + Zn(s)
Approximately zero, two solids produce two solids.
ii 2Pb(N03)2(s) -»2PbO(s) + 4N02(g) + 0z(g)
Significantly positive, a solid produces several moles of gases.
iii Mg0(s) + C02(g) -+ MgC03(s)
Significantly negative, a gas turns into a solid.
iv H20(l) — H20(g)
Significantly positive, a liquid turns into a gas.
b
i -7.8 J K-1 mol-1
ii +876.4 J K-1 mol-1
iii -174.8 J K-1 mol-1
iv +I 19 J K-1 mol-1
+493 kJ mol-1
ii 6000 K
-52 kJ mol-1
iii At which temperature is the reaction feasible?
It is feasible at 6000 K
b 5523 K
3
-284 J K-1 mol-1
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1 (a)
(i) (At 0 K) particles are stationary / not moving / not vibrating
(iiAs T increases, particles start to move / vibrate
(iii) Mark on temperature axis vertically below second ‘step’.
(iv)
· L2 corresponds to boiling / evaporating / condensing / l → g / g → l · And L1 corresponds to melting / freezing / · s →l / l →s · Bigger change in disorder for L2 / boiling · compared with L1 / melting |
(b)
(i
· ΔG = ΔH −TΔS · ΔH= c and (−)ΔS = m / ΔH and ΔS are constants (approx) |
(ii)
Because the entropy change / ΔS is positive / TΔS gets bigger |
(iii) Not feasible / unfeasible / not spontaneous
(c)
(i) + 44.5 J K-1 mol-1
(ii) At 5440 ΔH = TΔS
2
(a) ΔG = ΔH − TΔS
(b) 0.098 or 98
kJ K−1 mol-1 J K−1 mol-1
−ΔS/ΔS
(c) ΔG becomes negative
(d) Ammonia liquefies (so entropy data wrong/different)
3
(a) Enthalpy change when 1 mol of an (ionic) compound/lattice (under standard conditions) Is dissociated/broken/separated into its (component) ions
(b) There is an attractive force between the nucleus of an O atom and an external electron.
(c)
(d) LE MgO = 602 + 150 + 736 + 1450 + 248 − 142 + 844
= +3888 kJ mol-1
(e) Forms a protective layer/barrier of MgO / MgO prevents oxygen attacking Mg.
(f) ΔG = ΔH – TΔS
ΔS = (−602 − (−570)) × 1000 / 298
= −107 J K-1 mol-1 / −0.107 kJ K-1 mol-1
(g) 1 mol of solid and 0.5 mol of gas reactants form 1 mol solid products
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