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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

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} .

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(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} .

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