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3 years ago

15

Consider the reaction. At equilibrium, the concentrations of the different species are as follows. [NH3] = 0.105 M [N2] = 1.1 M

[H2] = 1.50 M What is the equilibrium constant for the reaction at this temperature?

1 answer:

Zina [86]3 years ago

4 0

Hey there!:

The reaction is as follows:

N2(g)+ 3 H2(g) ⇌ 2 NH3(g)

At equilibrium, the concentrations of the different species are as follows.:

[NH3] = 0.105 M

[N2] = 1.1 M

[H2] = 1.50 M

The equilibrium constant for the reaction is given as follows:

Keq = [NH3]² / [N2] [H2]³

Keq = (0.105)² / [1.1] [1.50]³

Keq = 0.00296 or 0.0030

The equilibrium constant for the reaction at this temperature is 0.0030.

Hope that helps!

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What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 3

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The value is $x = 0.0227 \ M$

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From the question we are told that

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The stability constant for $Ni(CN)_4 ^{2-}$ is $K_f = 1.0 *10^{31}$

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Generally the formation reaction for $Ni(CN)_4 ^{2-}$ is

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Combining both reaction we have

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Gnerally the equilibrium constant for this reaction is

$K_c = K_{sp} * K_f$

=> $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$

=> $K_c = 3.0*10^{12}$

Generally the I C E table for the above reaction is

$4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

initial [ I] 0.091 0 0

Change [C] -4x +x + x

Equilibrium [E ] 0.091 - 4x x x

Here is x is the amount in term of concentration that is lost by $CN^-$ and gained by $Ni(CN)_4 ^{2-}$ and $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

$K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=> $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=> $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=> $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

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Square both sides

$(119.8 - 5264x)^2 =x$

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So the molar solubility of nickel(II) sulfide at equilibrium is

$x = 0.0227 \ M$

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