You visited us 1 times! Enjoying our articles? Unlock Full Access!

Entropy

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 $A_{2}, B_{2}$ and ${A B}_{3}$ are $60, 40$ and $50 J K^{- 1} {m o l e}^{- 1}$ respectively. The above reaction will be in equilibrium at the temperature:

400 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500 K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

250 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

200 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $500 K$
For the reaction,

$1 / 2 A_{2} (g) + 3 / 2 B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 K J$

$Δ S_{A_{2}} = 60 J / K / m o l$

$Δ S_{B_{2}} = 40 J / K / m o l$

$Δ S_{{A B}_{3}} = 50 J / K / m o l$

$Δ S = \sum C_{i} Δ S_{p r o d u c t} - \sum C_{i} Δ S_{r e a c t a n t}$

$Δ S = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40) Δ S = - 40 J / K / m o l$

$Δ G = Δ H - T Δ S$

At equilibrium $Δ G = 0$

$∴ Δ H = T Δ S T = \frac{Δ H}{Δ S} = \frac{- 20 \times 1000}{- 40} T = 500 K$

Hence ,Option B is correct.

Suggest Corrections

Similar questions

\frac{1}{2} A_{2} (g) + \frac{3}{2} B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 k J

A_{2}, B_{2},

A B_{3}

60, 40,

50 J K^{- 1} m o l^{- 1}

1 / 2 A_{2} (g) + 3 / 2 B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 k J

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}

1 / 2 A_{2} (g) + 3 / 2 B_{2} (g) ⟶ A B_{3} (g); Δ H = - 20 k J

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}

\frac{1}{2} A_{2} (g) + \frac{3}{2} B_{2} (g) ⟶ {A B}_{3} (g);

Δ H = - 20 k J

A_{2}, B_{2}

{A B}_{3}

50 J . K^{- 1} . m o l^{- 1}

X_{2}, Y_{2} a n d X Y_{3}

50 J K^{- 1} m o l^{- 1}

\frac{1}{2} X_{2} + \frac{3}{2} Y_{2} ⇌ X Y_{3}; △ H = - 30 k J / m o l

Join BYJU'S Learning Program