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Equilibrium Constant and Standard Free Energy Change

For the equil...

Question

For the equilibrium :
$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$

$(G_{N_{2} O_{4}}^{o})_{298 K} = 100 k J m o l^{- 1} (G_{N O_{2}}^{o})_{298 K} = 50 k J m o l^{- 1}$

If 5 moles of each species is taken in a 1 litre container. Then, calculate the value of $Δ G$ at 298 K and predict the spontaneity of the reaction .

A

- 2.5 k J

and spontaneous.

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B

+ 2.5 k J

and non-spontaneous.

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C

+ 4 k J

and non-spontaneous.

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D

- 4 k J

and spontaneous.

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Solution

The correct option is C $+ 4 k J$ and non-spontaneous.
$C o n c e n t r a t i o n = \frac{M o l e s}{V o l u m e} [N_{2} O_{4}] = [N O_{2}] = \frac{5}{1} m o l L^{- 1} = 5 m o l L^{- 1}$

$\begin{matrix} N_{2} O_{4} & ⇌ & 2 N O_{2}; & G_{N_{2} O_{4}}^{0} = 100 k J m o l^{- 1} I n i t i a l l y : & 5 & 5 & G_{N O_{2}}^{0} = 50 k J m o l^{- 1} \end{matrix}$
$Δ G^{0} for reaction = 2 \times (G_{N O_{2}}^{0}) - G_{N_{2} O_{4}}^{0} = 2 \times 50 - 100 = 0$
so, $Δ G^{\circ} = 0$

$Δ G = △ G^{\circ} + 2.303 R T l o g Q$
$Q = \frac{[N_{2} O_{4}]^{2}}{[N O_{2}]}$
Now, $Δ G = 0 + 2.303 \times 8.314 \times 298 \times l o g \frac{[N O_{2}]^{2}}{[N_{2} O_{4}]} Δ G = 5705.85 \times l o g (\frac{5^{2}}{5}) Δ G = 5705.85 \times l o g (5) Δ G = 5707.85 \times 0.7 = 3994.01 J = + 3.99 k J \approx 4 k J$

Since, $Δ G > 0$ , the reaction is non spontaneous in forward direction.

Theory:

Gibbs free energy for reactions :

In case of chemical reactions.

$Δ G_{r e a c t i o n} = Σ G_{f} (p r o d u c t s) - Σ G_{f} (r e a c t a n t s)$

$Σ G_{f} (p r o d u c t s)$ is the sum of Gibbs free energies of formation of the products.

$Σ G_{f} (r e a c t a n t s)$ is the sum of Gibbs free energies of formation of the reactants.

In case of standard chemical reactions.

$Δ G_{r e a c t i o n}^{o} = Σ G_{f}^{o} (p r o d u c t s) - Σ G_{f}^{o} (r e a c t a n t s)$

$Σ G_{f}^{o} (p r o d u c t s)$ is the sum of standard Gibbs free energies of formation of the products.
$Σ G_{f}^{o} (r e a c t a n t s)$ is the sum of standard Gibbs free energies of formation of the reactants.

If $Δ G < 0$ Reaction is spontaneous
If $Δ G > 0$ Reaction is non - spontaneous
If $Δ G = 0$ then the reaction is at equilibrium.

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Similar questions

Q. For the equilibrium :

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

(G_{N_{2} O_{4}}^{o})_{298 K} = 100 k J m o l^{- 1} (G_{N O_{2}}^{o})_{298 K} = 50 k J m o l^{- 1}

If 5 moles of each species is taken in a 1 litre container. Then, calculate the value of

Δ G

at 298 K and predict the spontaneity of the reaction .

Q. For the equilibrium

N_{2} O_{4} ⇌ 2 N O_{2}

(G_{N_{2} O_{4}}^{o})_{298 K} = 100 k J m o l^{- 1} a n d (G_{N O_{2}}^{o})_{298 K} = 50 k J m o l^{- 1} .

When 5 mole/litre of each is taken, calculate the value of

△ G

k J m o l^{- 1}

for the reaction at 298 K.(nearest integer)

Q. For the equilibrium

N_{2} O_{4} ⇌ 2 N O_{2}

(G_{N_{2} O_{4}}^{o})_{298 K} = 100 k J m o l^{- 1} a n d (G_{N O_{2}}^{o})_{298 K} = 50 k J m o l^{- 1} .

When 5 mole/litre of each is taken, calculate the value of

△ G

k J m o l^{- 1}

for the reaction at 298 K.(nearest integer)

Q. For the equilibrium at

298

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); G_{N_{2} O_{4}}^{⊖} = 100 k J m o l^{- 1}

G_{N O_{2}}^{⊖} = 50 k J m o l^{- 1}

N_{2} O_{4}

N O_{2}

are taken initially in one litre container than which statement are correct

Q. Calculate the

Δ G

value from the following data and state whether the process will be spontaneous or non spontaneous at 298K. The heat of reaction is - 85.0 kJ and the entropy change for the reaction is +9.50 J/K.

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