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Entropy

Δ G0 for the ...

Question

$Δ G^{0}$ for the reaction $2 N O (g) + O_{2} (g) \to N_{2} O_{4} (g)$ at 298 k and 1 atm pressure, will be [given that the enthalpy of formation of NO (g) is $90.5 k J m o l^{- 1}$ .
The enthalpy of formation of $N_{2} O_{4} (g)$ is $9.7 k J m o l^{- 1}$ .
The standard entropy of NO (g) is $210 J k^{- 1}, O_{2} (g)$ is $205 J k^{- 1} m o l^{- 1}, N_{2} O_{4} (g) i s 304 J K^{- 1} c m o l^{- 1}$ ] :

A

- 52.92 k J

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B

- 75.64 k J

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C

- 64.24 k J

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D

- 83.27 k J

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Solution

The correct option is A $- 75.64 k J$
$\begin{matrix} D G = D X - T D S D X = D H_{f o r m} (p r o d u c t s) - D H_{r e a c t a n t s} \begin{matrix} \Rightarrow 9.7 K J | m o | - 2 \times 90.5 K 3 | m o | \Rightarrow - 171.3 K 3 | m o | \end{matrix} D S = S_{p r o d u c t} - S_{r e a c t a n t} \begin{matrix} \Rightarrow 304 - 205 - 2 \times 210 \Rightarrow 304 - 205 - 420 \Rightarrow - 321 J K^{- 1} | m o | \end{matrix} D 4 = - 171.3 k J | m o | + 29 + 321 J K^{- 1} | m o | \Rightarrow - 75.64 k J | m o | \end{matrix}$

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