(1)www.pianetachimica.it 43 IChO - Soluzioni preliminari dei Problemi Preparatori 4 Problema 4 Magnesium compounds a) i) 2 Mg(s

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43 IChO - Soluzioni preliminari dei Problemi Preparatori 4

Problema 4 Magnesium compounds a) i) 2 Mg(s) + O2(g) → 2 MgO(s)

ii) 2 Mg_(s) + CO_2(g) → 2 MgO(s) + C_(s) b) i) Mg(s) + 2 H2O(l) → 2 Mg(OH)2(s) + H2(g)

ii) MgO_(s) + H₂O_(l) → 2 Mg(OH)_2(s)

c) A=Mg3N2 B=NH3 C=N2H4 D=NaNH2 E=NaN3

A) 3 Mg(s) + N2(g) → Mg3N2(s)

B) Mg₃N_2(s) + 6 H₂O_(l) → 2 NH_3(g) + 3 Mg(OH)_2(aq) C) 2 NH3(g) + ClO^-(aq) → Cl^-(aq) + N2H4(aq) + H2O(l)

2 NH3(g) + H2O2(aq) → N2H4(aq) + 2 H2O(l)

D) 2 NH_3(g) + 2 Na_(s) → 2 NaNH_2(s) + H_2(g)

E) 2 NaNH2(s) + N2O(g) → NH3(g) + NaOH(s) + NaN3(s)

2 NaN3(s) → 2 Na(s) + 3N2(g)

d)

e) 3 N₂H_4(l) → N_2(g) + 4 NH_3(g)

f) 3 N₂H_4(l) → 3 N₂H_4(g) → N_2(g) + 4 NH_3(g)

ΔE°reaz. = ((3·95,4 - 3·50,6)+(12·389 + 3·163 - (946 + 12·389))) kJ/mol = - 322,6 kJ/mol Usando la legge di Hess

3/4 N_2(g) + 3/2 H_2(g) → 3/4 N2H_4(l) ∆H1 = 3/4 ∆H°f, idrazina

3/4 N₂H_4(l) → 1/4 N2(g) + NH_3(g) ∆H2 = 1/4 ΔE°_reaz.

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1/2 N_2(g) + 3/2 H_2(g) → NH3(g) ∆H°f, ammoniaca = ∆H₁ + ∆H₂ = - 42,7 kJ/mol

g) mol

mol g

mL mL g

MM d n V

H N

H N H N H

N 0,0627

052 , 32

0045 , 1 00 , 2

4 2

4 2 4 2 4

2 ⋅ =

⋅ =

=

mol n

n_{tot NH} 0,1044 3

5

4

2 =

=

atm m Pa

K K mol mol J

V T R

p n^tot 2,59 10 2,55

10 00 , 1

298 314

, 8 1044 , 0

5 3

3 = ⋅ =

⋅

⋅ ⋅

= ⋅

⋅

= ⋅ ₋

h) J

atm K atm

K mol mol J

p T p R n w

f i tot

isoterma 242

1 55 , ln2 298 314

, 8 1044 , 0

ln ⋅ ⋅ =−

⋅ ⋅

−

=

⋅

⋅

⋅

−

=

Soluzione proposta da

Daniela Lorizio e Ivan Palazzo ITIS Dell’Erba – Castellana Grotte