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For the reaction, $X_2O_{4(l)} \to 2XO_{2(g)}$

$\Delta U$ = 2.1 kcal, $ΔS$ = 20 cal K-1 at 300 K

Hence, $ΔG$ is

(a) 2.7 kcal

(b) - 2.7 kcal

(c) 9.3 kcal

(d) - 9.3 kcal

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Ans: (b) - 2.7 kcal

Sol: $ΔH = ΔU + Δn_g RT$

Given, $ΔU = 2.1 kcal, Δn_g = 2,$

$R = 2 × 10^{-3} kcal, T = 300 K$

$∴ ΔH = 2.1 + 2 × 2 × 10^{-3} × 300 = 3.3 kcal$

Again, $ΔG = ΔH - TΔS$

Given, $ΔS = 20 × 10^{-3} kcal K^{-1}$

On putting the values of $ΔH$ and $ΔS$ in the equation,

we get

$ΔG = 3.3 - 300 × 20 × 10^{-3}$

$= 3.3 - 6 × 10^3 × 10^{-3} = - 2.7 kcal$

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