The equilibrium constant for the reaction- AgCls- Agaq-Claq- is-Given that - Gf AgCl0-109-7kJ- - Gf Ag-0-77-1kJ- - Gf Cl-0-131-2kJ

The correct option is B 2× 10^ 10 As we know, at equlibrium #160; ΔG^0 for the reaction =77.1 +( 131.2) ( 109.7)=55.6kJ ΔG^0= 2.303 RTlog ...

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Standard XII

Physics

Calorimetry

The equilibri...

Question

The equilibrium constant for the reaction, $A g C l_{(s)} ⇌ A g_{(a q)}^{+} + C l_{(a q)}^{-}$ is:

(Given that $: Δ G_{f A g C l}^{0} = - 109.7 k J$ ; $Δ G_{f A g^{+}}^{0} = 77.1 k J$ ; $Δ G_{f C l^{-}}^{0} = - 131.2 k J$ )

2 \times 10^{- 10}

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6 \times 10^{- 10}

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8 \times 10^{- 10}

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2 \times 10^{- 8}

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Solution

The correct option is B $2 \times 10^{- 10}$
As we know, at equlibrium $Δ G^{0}$ for the reaction $= 77.1 + (- 131.2) - (- 109.7) = 55.6 k J$
$Δ G^{0} = - 2.303 R T log K$
$log K = \frac{- Δ G^{0}}{2.303 R T} = \frac{- 55.6 \times 10^{3}}{2.303 \times 8.314 \times 298} = - 9.75$

$K = K_{e q} = 1.8 \times 10^{- 10} ≃ 2 \times 10^{- 10}$

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Similar questions

Q. The equilibrium constant for the reaction

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

x \times 10^{- 10}

. Then, the value of

x

is :
(Given

: Δ G_{f (A g C l)}^{0} = - 109.7 k J;

Δ G_{f (A g^{+})}^{0} = 77.1 k J;

Δ G_{f (C l^{-})}^{0} = - 131.2 k J

)

Q. Using the following data, calculate equilibrium constant K for the reaction

AgC l_{(s)} ⇌ {Ag}_{(a q)}^{+} + {Cl}_{(a q)}^{-}

Δ G_{f (AgCl)}^{\circ} = - 109.7 kJ, Δ G_{f (A g^{+})}^{\circ} = 77.1 kJ,

Δ G_{f (C l^{-})}^{\circ} = - 131.2 kJ,

Q. For the reaction,

A g_{(aq)}^{+} + C l_{(aq)}^{-} ⇌ A g C l_{(s)}

Given:

\begin{matrix} Species & Δ G_{f}^{\circ} (kJ/mol) A g_{(aq)}^{+} & + 77 C l_{(aq)}^{-} & - 129 A g C l_{(s)} & - 109 \end{matrix}

Calculate

E_{cell}^{\circ}

298 K

.
Also find the solubility product of

A g C l

Q. Write the number for the given expanded form
2*10^6 + 3*10^4 + 5*10^2 + 8*10^0

Write the numeral whose expanded form is given below:
(i) $6 \times 10^{4} + 3 \times 10^{3} + 0 \times 10^{2} + 7 \times 10^{1} + 8 \times 10^{0}$
(ii) $9 \times 10^{6} + 7 \times 10^{5} + 0 \times 10^{4} + 3 \times 10^{3} + 4 \times 10^{2} + 6 \times 10^{1} + 2 \times 10^{0}$
(iii) $8 \times 10^{5} + 6 \times 10^{4} + 4 \times 10^{3} + 2 \times 10^{2} + 9 \times 10^{1} + 6 \times 10^{0}$

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The equilibrium constant for the reaction, AgCl(s)⇌Ag+(aq)+Cl−(aq) is:(Given that :ΔG0fAgCl=−109.7kJ; ΔG0fAg+=77.1kJ; ΔG0fCl−=−131.2kJ )

The correct option is B 2×10−10As we know, at equlibrium ΔG0 for the reaction =77.1+(−131.2)−(−109.7)=55.6kJΔG0=−2.303RTlogKlogK=−ΔG02.303RT=−55.6×1032.303×8.314×298=−9.75K=Keq=1.8×10−10≃2×10−10

The equilibrium constant for the reaction, $A g C l_{(s)} ⇌ A g_{(a q)}^{+} + C l_{(a q)}^{-}$ is:

(Given that $: Δ G_{f A g C l}^{0} = - 109.7 k J$ ; $Δ G_{f A g^{+}}^{0} = 77.1 k J$ ; $Δ G_{f C l^{-}}^{0} = - 131.2 k J$ )