Calculate a - G- and b the equilibrium constant for the formation of NO 2 from NO and O 2 at 298 K NO g -1 - 2 O 2 g - NO 2 g where -f G- NO 2-52-0 kJ - mol-f G- NO -87-0 kJ - mol-f G- O 2-0 kJ - mol

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Standard XII

Chemistry

Reaction Quotient

Calculate a Δ...

Question

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

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Solution

(a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 + 0}

= – 35.0 kJ mol^–1

(b) We know that,

ΔG°= RT log K_c

ΔG°= 2.303 RT log K_c

Hence, the equilibrium constant for the given reaction K_c is 1.36 × 10⁶

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

Q. What are the value of

△ G^{\circ}

and the equilibrium constant for the formation of

N O_{2}

from

N O

and

O_{2}

at 298 K. Given log

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

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

where

△ G^{\circ} (N O_{2}) = 52.0

kJ/mol

△ G^{\circ} (N O) = 87.0

kJ/mol

△ G^{\circ} (O_{2}) = 0

kJ/mol

Q. Calculate

Δ G^{0}

and the equilibrium constant for the formation of

N O_{2}

from NO and

O_{2}

at 298K

N O + 1 / 2 O_{2} ⇌ N O_{2}

, where

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

Q. What are the value of

△ G^{\circ}

and the equilibrium constant for the formation of

N O_{2}

from

N O

and

O_{2}

at 298 K. Given log

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

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

where

△ G^{\circ} (N O_{2}) = 52.0

kJ/mol

△ G^{\circ} (N O) = 87.0

kJ/mol

△ G^{\circ} (O_{2}) = 0

kJ/mol

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Calculate a) ΔG°and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298 K where ΔfG° (NO2) = 52.0 kJ/mol ΔfG° (NO) = 87.0 kJ/mol ΔfG° (O2) = 0 kJ/mol

(a) For the given reaction, ΔG° = ΔG°( Products) – ΔG°( Reactants) ΔG° = 52.0 – {87.0 + 0} = – 35.0 kJ mol–1 (b) We know that, ΔG° = RT log Kc ΔG° = 2.303 RT log Kc Hence, the equilibrium constant for the given reaction Kc is 1.36 × 106

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

(a) For the given reaction,

ΔG° = ΔG°( Products) – ΔG°( Reactants)

ΔG° = 52.0 – {87.0 + 0}

= – 35.0 kJ mol^–1

(b) We know that,

ΔG°= RT log K_c

ΔG°= 2.303 RT log K_c

Hence, the equilibrium constant for the given reaction K_c is 1.36 × 10⁶