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Byju's Answer

Standard XII

Chemistry

Law of Mass Action

How can be eq...

Question

How can be equilibrium concentration of $A g^{+}$ can be reduced in a solution of $N a^{+}, C l^{+}, A g^{+}$ and $N O_{3}$ , in contact with solid $A g C l$ . $N a^{+} (a q) + C l (a q) + A g^{+} (a q) + N O_{3} (a q) ⇌ A g C l (s) + N a^{+} (a q) + N O_{3} (a q) Δ H = 65.9 k J$ ?

Add

N a C l

in the solution.

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Some other salt that produces

C l^{+}

in the solution.

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Cool the solution

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All of the above

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Solution

The correct option is D All of the above
$N a^{+} (a q) + C l^{(} a q) + A g^{+} (a q) + N O_{3}^{(} a q) ⇌ A g C l (s) + N a^{+} (a q) + N O_{3}^{(} a q) Δ H = 65.9 k J$
As we know,
To increase product, we have to add reactant (or common ion) so by adding $C l^{+}$ $(N a C l)$ product increases.
Also as this is an exothermic reaction so it will be favoured by losing heat.

Suggest Corrections

Similar questions

Q. ways in which the equilibrium concentration of

A g^{+}

can be reduced in a solution of

N a^{+}

C l^{-}

A g^{+}

and

N O_{3}^{-}

, in contact with solid

A g C l .

N a^{+} (a q) + C l^{-} (a q) + A g^{+} (a q) + N O_{3}^{-} (a q) ⇌ A g C l (s) + N a^{+} (a q) + N O_{3}^{-} (a q) Δ H = - 65.9 k J

(I) For the reaction,

A g^{+} (a q) + C l^{-} (a q)_{\to} A g C l (s)

Write the cell representation of above reaction and calculate

E^{o}

at 298 K.

Given:

Species	$Δ G_{f} (k J / m o l)$
$A g^{+} (a q)$	+77
$C l^{-} (a q)$	-129
$A g C l (s)$	-109

(II) Calculate the

{log}_{10} K_{s p} (A g C l)

at 298 K.

(III) If

6.539 \times 10^{- 2} g

of metallic zinc is added to 100 ml saturated solution of

A g C l

, find the value of

{log}_{10} \frac{[Z n^{2 +}]}{[A g^{+}]^{2}}

(IV)How many moles of Ag will be precipitated in the above reaction?

Given that-

A g^{+} + e^{-} \to

Ag;

E^{o} = 0.80 V

Z n^{2 +} + 2 e^{-} \to

Zn;

E^{o} = - 0.76 V

(It was given that Atomic mass of Zn

= 65.39

)

Q. For the reaction

A g^{+} (a q) + C l^{-} (a q) ⇌ A g C l (s)

; the

Δ G^{o}

values for

A g^{+} (a q), C l^{-} (a q)

and

A g C l (s)

are + 77, -129 and -109 kJ mol

^{- 1}

. Write the cell representation of above reaction and calculate

K_{s p}

of AgCl at 298 K.
(Multiply the ans by

10^{10}

)

Q. calculate

Δ G_{f}^{⊖}

of the reaction :

A g^{\oplus} (a q) + C l^{⊖} (a q) \to A g C l (s)

Given :

Δ G^{⊖_{A g C l}} = - 109 k J m o l^{- 1}

Δ G_{(C l^{⊖})}^{⊖} = - 129 k J m o l^{- 1}

Δ G_{(A g^{\oplus)}}^{⊖} = - 77 k J m o l^{- 1}

Q. The reaction,

\frac{1}{2} H_{2} (g) + A g C l (s)

\to

H^{+} (a q .) + C l^{-} (a q .) + A g (s)

occurs in the galvanic cell:

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How can be equilibrium concentration of Ag+ can be reduced in a solution of Na+,Cl+,Ag+ and NO3, in contact with solid AgCl. Na+(aq)+Cl(aq)+Ag+(aq)+NO3(aq)⇌AgCl(s)+Na+(aq)+NO3(aq)ΔH=65.9kJ ?

The correct option is D All of the aboveNa+(aq)+Cl(aq)+Ag+(aq)+NO(3aq)⇌AgCl(s)+Na+(aq)+NO(3aq)ΔH=65.9kJAs we know,To increase product, we have to add reactant (or common ion) so by adding Cl+ (NaCl) product increases.Also as this is an exothermic reaction so it will be favoured by losing heat.

How can be equilibrium concentration of $A g^{+}$ can be reduced in a solution of $N a^{+}, C l^{+}, A g^{+}$ and $N O_{3}$ , in contact with solid $A g C l$ . $N a^{+} (a q) + C l (a q) + A g^{+} (a q) + N O_{3} (a q) ⇌ A g C l (s) + N a^{+} (a q) + N O_{3} (a q) Δ H = 65.9 k J$ ?