Consider this equilibrium, for which $Δ H < 0$ .
$H g O (s) + 4 I^{-} (a q .) + H_{2} O (l) ⇌ {H g I_{4}}^{2 -} + 2 {O H}^{-}$
Which changes will increase the equilibrium concentration ${H g I_{4}}^{2 -}$ ?
$I$ . Increasing the mass of $H g O (s)$ present
$I I$ . Increasing $[I^{-}]$
$I I I$ . Adding $1 M$ $H C l$

I

only

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I I

only

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I I

and

I I I

only

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I, I I

and

I I I

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Solution

The correct option is C $I I$ and $I I I$ only
We have to increase the con. of $H g I_{4}$ then according to le chandelier principal
1- increase the con. of HgO is wrong because Hgo is solid then no. the effect in equ.
2 - increase the con. of iodine ion
3 - and adding 1 M of HCL because it's will decrease the con. of OH ion on neutralizing it.

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Similar questions

H g O (s) + 4 I^{-} (a q) + H_{2} O (l) ⇌ H g I_{4}^{2 -} (a q) + 20 H^{-} (a q)

H g I_{4}^{2 -}

f (x) = a^{x} \forall ϵ R

Which of the following graphs represents exothermic reaction ?

(a) Only I
(b) Only II
(c) Only III
(d) I and II

O_{2}^{+}

N O

N_{2}^{+}

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