For the reaction given below the values of standard Gibbs free energy of formation at 298 K are given. What is the nature of reaction?
$I_{2} + H_{2} S \to 2 H I + S$
$△ G_{f}^{o} (H I) = 1.8 k J m o l^{- 1}, △ G_{f}^{o} (H_{2} S) = 33.8 k J m o l^{- 1}$

Non-spontaneous in forward direction.

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

Spontaneous in forward direction.

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

Spontaneous in backward direction.

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

Non-spontaneous in both forward and backward directions.

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

Open in App

Solution

The correct option is B Spontaneous in forward direction.
$I_{2} + H_{2} S \to 2 H I + S$
$△ G^{o} = \sum G_{f (P r o d u c t s)}^{o} - \sum G_{f (R e a c t a n t s)}^{o}$
$△ G^{o} = (2 \times 1.8 + 0) - (0 + 33.8) = - 30.2 k J$
Hence, reaction is spontaneous in forward direction.

Suggest Corrections

Similar questions

I_{2} + H_{2} S \to 2 H I + S

△ G_{f}^{o} (H I) = 1.8 k J m o l^{- 1}, △ G_{f}^{o} (H_{2} S) = 33.8 k J m o l^{- 1}

I_{2} + H_{2} S \to 2 H I + S

Δ G_{f}^{0} (H I) = 1.8 k J m o l^{- 1}

Δ G_{f}^{0} (H_{2} S) = 33.8 k J m o l^{- 1}

I_{2} + H_{2} S \to 2 H I + S

△ G_{f}^{o} (H I) = 1.8 k J m o l^{- 1}, △ G_{f}^{o} (H_{2} S) = 33.8 k J m o l^{- 1}

^{- 1}

^{- 1}

^{- 1}

K J m o l^{- 1} a t 298 K

Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s), E^{\circ} = 2 V a t 298 K

(F a r a d a y^{'} s c o n s t a n t, F = 96000 C m o l^{- 1})

Join BYJU'S Learning Program