For the reaction, , cal at 300 K Hence, is, Thermodynamics, Spontaneity [2014]

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Solution

Q.
For the reaction, $X_{2} O_{4 (l)} ⟶ 2 X O_{2 (g)}$

$Δ U = 2.1$ $kcal$ , $Δ S = 20$ cal $K^{- 1}$ at 300 K

Hence, $Δ G$ is [2014]

1 2.7 kcal

2 - 2.7 kcal

3 9.3 kcal

4 - 9.3 kcal

Ans.
(2)

$Δ H = Δ U + Δ n_{g} R T$

Given, $Δ U = 2.1$ kcal, $Δ n_{g} = 2$ ,

$R = 2 \times 10^{- 3}$ kcal, T = 300 K

$∴$ $Δ H = 2.1 + 2 \times 2 \times 10^{- 3} \times 300 = 3.3$ kcal

Again, $Δ G = Δ H - T Δ S$

Given, $Δ S = 20 \times 10^{- 3}$ kcal $K^{- 1}$

On putting the values of $Δ H$ and $Δ S$ in the equation, we get

$Δ G = 3.3 - 300 \times 20 \times 10^{- 3} = 3.3 - 6 \times 10^{3} \times 10^{- 3} = - 2.7$ kcal

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