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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For the equil...

Question

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 $△ G$ in $k J m o l^{- 1}$ for the reaction at 298 K.(nearest integer)

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Solution

$\begin{matrix} N_{2} O_{4} & ⇌ & 2 N O_{2}; & G_{N_{2} O_{4}}^{0} = 100 k J m o l^{- 1} c o n c . a t t = 0 & 5 & 5 & G_{N O_{2}}^{0} = 50 k J m o l^{- 1} \end{matrix}$

$△ G^{0}$ for reaction $= (2 \times G^{0} N O_{2}) - G_{N_{2} O_{4}}^{0} = (2 \times 50) - 100 = 0$
$△ G = △ G^{0} + 2.303 R T l o g Q$
Now, $△ G = 0 + 2.303 \times 8.314 \times 10^{- 3} \times 298 l o g \frac{5^{2}}{5}$
$= + 3.988 k J m o l^{- 1} \approx 4 k J m o l^{- 1}$

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Similar questions

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}

and

G_{N O_{2}}^{⊖} = 50 k J m o l^{- 1}

. If 5 mol of

N_{2} O_{4}

and 2 moles of

N O_{2}

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

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} (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 .

The reaction N $_{2}$ O $_{4}$ (g) $⇌$ 2NO $_{2}$ (g) is
carried out at 298 K and 20 bar. Five moles of each of N $_{2}$ O $_{4}$ and
NO $_{2}$ , are taken initially. Given: $Δ$ $_{f}$ G $^{\circ}$
(N $_{2}$ O $_{4}$ ) = 100 kJ mol $^{- 1}$ and $Δ$ $_{f}$ G $^{\circ}$
(NO $_{2}$ ) = 50 kJ mol $^{- 1}$ . The reaction proceeds at an initial pressure
of 20 bar. Determine the direction in which the reaction proceeds to achieve
equilibrium. Also determine the amounts of N $_{2}$ O $_{4}$ and NO $_{2}$ when the reaction attains equilibrium.

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Gibbs Free Energy & Spontaneity

Standard XII Chemistry

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For the equilibrium N2O4⇌2NO2 (GoN2O4)298K=100kJmol−1 and (GoNO2)298K=50kJmol−1. When 5 mole/litre of each is taken, calculate the value of △G in kJ mol−1 for the reaction at 298 K.(nearest integer)

N2O4⇌2NO2;G0N2O4=100kJ mol−1conc.at t=055G0NO2=50kJ mol−1 △G0 for reaction =(2×G0NO2)−G0N2O4=(2×50)−100=0 △G=△G0+2.303RTlogQ Now, △G=0+2.303×8.314×10−3×298 log525 =+3.988 kJ mol−1≈4 kJ mol−1

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 $△ G$ in $k J m o l^{- 1}$ for the reaction at 298 K.(nearest integer)