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

Introduction

Given the equ...

Question

Given the equilibrium constants:- $C I^{-} + H g C I^{+} ⇌ H g C I_{2}$ ; $K_{1} = 3 \times 10^{6}$ .
$H g C I_{2} + C I^{-} ⇌ H g C I_{3}^{-}$ ; $K_{2} = 8.9$ .
The equilibrium constants for the disproportionation equilibrium $2 H g C I_{2} ⇌ H g C I^{+} H g C I_{3}^{-}$ is :

3 \times 10^{- 6}

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

4 \times 10^{- 6}

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

3.5 \times 10^{- 6}

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

4.5 \times 10^{- 6}

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

Open in App

Solution

The correct option is A $3 \times 10^{- 6}$
$C I^{-} + H g C I^{+} ⇌ H g C I_{2}$ ; $K_{1} = 3 \times 10^{6}$ ...... (1)
$H g C I_{2} + C I^{-} ⇌ H g C I_{3}^{-}$ ; $K_{2} = 8.9$ ......(2)
$2 H g C I_{2} ⇌ H g C I^{+} H g C I_{3}^{-}$ ......(3).
The reaction (1) is reversed and added to reaction (2) to obtain the reaction (3).
Hence, the equilibrium constant for the third reaction will be $K = \frac{8.9}{3 \times 10^{6}} = 3 \times 10^{- 6}$

Suggest Corrections

Similar questions

H g C l^{\oplus} + C l^{⊖} \to H g C l_{2}, K_{1} = 3 \times 10^{6}

H g C l_{2} + C l^{⊖} \to H g C l_{3}^{⊖}, K_{2} = 8.9

2 H g C l_{2} \to H g C l^{\oplus} + H g C l_{3}^{⊖}

K_{1} = 10^{6}, K_{2} = 10^{- 4}, K_{3} = 10, K_{4} = 1

A g^{+} + N H_{3}

⇋

[A g (N H_{3})]^{+}

K_{1} = 3.5 \times 10^{- 3}

[A g (N H_{3})]^{+} + N H_{3}

⇋

[A g (N H_{3})_{2}]^{+}

K_{2} = 1.7 \times 10^{- 3}

[A g (N H_{3})_{2}]^{+}

2 H_{2} O (g) ⇋ 2 H_{2} (g) + O_{2} (g); K_{1} = 6.4 \times 10^{- 8}

2 C O_{2} (g) ⇋ 2 C O (g) + O_{2} (g); K_{2} = 1.6 \times 10^{- 6}

H_{2} (g) + C O_{2} (g) ⇋ C O (g) + H_{2} O (g)

H g C l_{2}

H g C l^{+}

H g C l_{3}^{-}

H g C l^{+} + C l^{-} ⇌ H g C l_{2}; K_{1} = 3 \times 10^{6}

H g C l_{2} + C l^{-} ⇌ H g C l_{3}^{-}; K_{2} = 9.0

Join BYJU'S Learning Program