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Calorimetry

For hypotheti...

Question

For hypothetical reversible reaction $1 / 2 A_{2} (g) + 3 / 2 B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 k J$ if standard entropies of $A_{2}, B_{2}, a n d A B_{3} a r e 60, 40, a n d 50 J K^{- 1} m o l^{- 1}$ , respectively. What will be the temperature(in K) in the given reaction?

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Solution

As we know,
At equilibrium, $Δ S_{t o t a l} = 0$
$∴ S_{P}^{⊖} - S_{R}^{⊖} = 0$
Also,
$Δ G = Δ H - T Δ S = 0$ at eqm.
so,
$- 20, 000 = T (50 - 30 - 60)$
$T = 500 K$

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