Standard entropy of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60,40 and $50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. For the reaction, $\frac{1}{2} X_{2}+\frac{3}{2} Y_{2} \rightarrow X Y_{3}, \Delta H=-30 \mathrm{~kJ}$, to be at equilibrium, the temperature will be

JEE Main · 2008 · 27 Apr 2008 · Offline · Chemical Thermodynamics

Standard entropy of

X_{2}, Y_{2}

and

X Y_{3}

are 60,40 and

50 JK^{- 1} mol^{- 1}

, respectively. For the reaction,

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} \to X Y_{3}, Δ H = - 30 kJ

, to be at equilibrium, the temperature will be

Answer: (C)

First find the reaction entropy,

Δ S = S_{X Y_{3}} - (\frac{1}{2} S_{X_{2}} + \frac{3}{2} S_{Y_{2}}) = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40) = 50 - 90 = - 40 J K^{- 1} mo l^{- 1}

. At equilibrium the free energy change is zero, so

Δ H = T Δ S

. With

Δ H = - 30000 J

and

Δ S = - 40 J K^{- 1}

, the temperature is

T = \frac{- 30 000}{- 40} = 750 K

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option