The standard entropies of A₂,B₂ and AB₃ are 60, 40 and 50 JK^-1 mol^-1 respectively. The following reaction would be an equilibrium reaction at the temperature …K
$\frac{1}{2} A_{2} + \frac{3}{2} B_{2} ⇌ {AB}_{3} + 30 .0 kJ$

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1250 K

750 K

1000 K

500 K

answer is B.

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Detailed Solution

At equilibrium

$ΔG = ΔH - TΔS = 0 or ΔH = TΔS$

$Δ H = - 30.0 k J$

${ΔS}_{r} = S_{{AB}_{3}} - (\frac{1}{2} S_{A_{2}} + \frac{3}{2} S_{B_{2}})$ =50 $- (\frac{1}{2} \times 60 + \frac{3}{2} \times 40)$

= 50 – (30 + 60) = -40 JK^-1

$∴ T = \frac{ΔH}{ΔS} = \frac{30000}{40} = 750$ K

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