An aqueous solution of volume $500$ ml contains the reaction
$2 A g^{+} (a q .) + C u (s) ⇌ C u^{2 +} (a q .) + 2 A g (s)$ in equilibrium with $[C u^{2 +} (a q)] =$ $x M$ . Now, $500$ ml of water is further added. On reset of above equilibrium $[C u^{2 +} (a q)]$ will be:

x M

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2 x M

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between

x M

and

\frac{x}{2} M

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less than

\frac{x}{2} M

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Solution

The correct option is D less than $\frac{x}{2} M$
The equilibrium reaction is $2 A g^{+} (a q .) + C u (s) ⇌ C u^{2 +} (a q .) + 2 A g (s)$ .

Let $y$ and $x$ be the equilibrium concentrations of $A g^{+}$ and $C u^{2 +}$ respectively.

The expression for the equilibrium concentration is $K_{c} = \frac{x}{y^{2}}$ .

When the solution is diluted, the concentrations of $A g^{+}$ and $C u^{2 +}$ will be halved.

$0.5 y$ and $0.5 x$ will be the new concentrations of $A g^{+}$ and $C u^{2 +}$ respectively.

The expression for the reaction quotient will be

$Q_{c} = \frac{x / 2}{(y / 2)^{2}}$
The value of the reaction quotient is greater than the value of the equilibrium constant $(Q_{c} > K_{c})$ .
Hence, the reaction will shift in the backward direction.

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Similar questions

500 m L

2 A g^{+} + C u ⇌ C u^{2 +} + 2 A g

[C u^{2 +}]

x M

500 m L

[C u^{2 +}]

C u (s) + 2 {A g}^{+} (a q) \to {C u}^{2 +} (A q) + 2 A g (s)

(K)

E_{c e l l}^{\circ} = 0.46 V

C_{u} (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s); E^{0} = 0.46 V a t 298 K

Calculate the equilibrium constant of the reaction.
$C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s)$ , $E_{c e l l}^{⊖} = 0.46 V$

C u (s) + 2 A g^{+} (a q) ⇋ C u^{2 +} (a q) + 2 A g (s),

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