Question

4. From the following data:
C(graphite) → CgAH = 716.7 kJ/mole
Cadiamond) → Ce)
AH = 714.8 kJ/mole
Calculate AH for the reaction C(graphtie)
→ C(diamond).
(A) 714.8 kJ
(B) -1.9 kJ
(C) Zero
(D) 1.9 kJ
If heat ic bo​

Answer 1

Answer:

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For the given reaction,

NO (g) + 1/2 O2 (g) ⇌ NO2 (g)

Given:

Standard free energy change of the formation of the product (NO2)

⇒ΔfG° (NO2) = 52.0 kJ/mol

Standard free energy change of the formation of one reactant (NO)

⇒ΔfG° (NO) = 87.0 kJ/mol

Standard free energy change of the formation of second reactant(O2)

⇒ΔfG° (O2) = 0 kJ/mol

a) Calculation of ΔG°

ΔG° = Difference in free energy of the reaction when all the reactants and products are in the standard state (1 atmospheric pressure and 298K) Hence,

ΔG° = ∑ ΔfG° (Products) - ∑ ΔfG° (Reactants)

⇒ΔG° = ΔfG° (NO2) – [ΔfG° (NO) + 1/2 ΔfG° (O2)]

⇒ΔG° = 52.0 – [87.0 + 1/2 x 0]

⇒ΔG° = -35.0 kJ mol-1

b) Equilibrium constant for the formation of NO2 from NO and O2 at 298K

From standard free change of a reaction, we know that,

ΔG° = -2.303 RT log Kc

Where, ΔG° is the standard free change of a reaction, value of ΔG° is -35000 (calculated in part(a))

R = gas constant, value of R is 8.314 J/mol-K

T= Absolute temperature, value of T is 298K (given)

Kc= Equilibrium constant

Using the formula, we write,

log Kc =

⇒log Kc =

⇒log Kc = 6.134

∴ Kc = antilog (6.134)

⇒Kc = 1.361 x 106

Answer 2

Answer:

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Explanation:

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