Calculate-a - G0-b the equilibrium constant for the formation of NO2 from NO and O2 at 298 K- NOg-12O2g - NO2g where - f G0 NO2-52-0 kJ-mol - f G0NO-87-0 kJ-mol - f G0O2-0 kJ-mol-

(a) Δ G^0 = Δ_f G^0 (NO_2) [Δ_f G^0(NO)+ 12Δ_f G^0(O_2)] = 52.0 87.0 12 × 0 = 35 kJ/mol(b) log K = Δ G^0/2.303RT = 35 × 10^3/2.303 × 8.314 ...

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Kohlrausch Law

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Question

Calculate:
a) $Δ G^{0}$ .
b) the equilibrium constant for the formation of $N O_{2}$ from NO and $O_{2}$ at 298 K.

$N O (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)$

where $Δ_{f} G^{0} (N O_{2}) = 52.0$ kJ/mol $Δ_{f} G^{0} (N O) = 87.0$ kJ/mol $Δ_{f} G^{0} (O_{2}) = 0$ kJ/mol.

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Solution

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△ G^{\circ}

N O_{2}

N O

O_{2}

(1.365 \times 10^{6}) = 6.14

N O (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)

△ G^{\circ} (N O_{2}) = 52.0

△ G^{\circ} (N O) = 87.0

△ G^{\circ} (O_{2}) = 0

Δ G^{0}

N O_{2}

O_{2}

N O + 1 / 2 O_{2} ⇌ N O_{2}

Δ_{f} G^{0} (N O_{2}) = 52.0 k J / m o l

Δ_{f} G^{0} (N O) = 87.0 k J / m o l

Δ_{f} G^{0} (O_{2}) = 0 k J / m o l

Calculate a) ΔG°and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

where Δ_fG° (NO₂) = 52.0 kJ/mol

Δ_fG° (NO) = 87.0 kJ/mol

Δ_fG° (O₂) = 0 kJ/mol

[[Q]]

Calculate a) G°and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K

where _fG° (NO₂) = 52.0 kJ/mol

_fG° (NO) = 87.0 kJ/mol

_fG° (O₂) = 0 kJ/mol

Calculate

$Δ G^{0}$

the equilibrium constant for the formation of $N O_{2}$ from NO and $O_{2}$ at 298 K
$N O (g) + \frac{1}{2} O_{2} (g) \leftrightarrow N O_{2} (g)$

Where:
$Δ_{f} G^{\circ} (N O_{2}) = 52.0 k J / m o l Δ_{f} G^{\circ} (N O) = 87.0 k J / m o l Δ_{f} G^{\circ} (O_{2}) = 0 k J / m o l$

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