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kirill115 [55]
3 years ago
10

Consider the reaction. Upper H subscript 2 upper o (g) plus upper C l subscript 2 upper O (g) double-headed arrow 2 upper H uppe

r C l upper O (g). At equilibrium, the concentrations of the different species are as follows. [H2O] = 0.077 M [Cl2O] = 0.077 M [HClO] = 0.023 M What is the equilibrium constant for the reaction at this temperature?
Chemistry
2 answers:
Vlad [161]3 years ago
6 0

Answer:

0.089

Explanation:

Step 1:

The balanced equation for the reaction is given below:

H2O + Cl2O <=> 2HClO

Step 2:

Data obtained from the question. This includes:

Concentration of H2O, [H2O] = 0.077 M

Concentration of Cl2O, [Cl2O] = 0.077 M

Concentration of HClO, [HClO] = 0.023 M

Equilibrium constant, K =?

Step 3:

Determination of the equilibrium constant. This is illustrated below:

The equilibrium constant for the above reaction is given below:

K = [HClO]^2 / [H2O] [Cl2O]

K = (0.023)^2 / (0.077 x 0.077)

K = 0.089

Therefore, the equilibrium constant for the above reaction is 0.089

dedylja [7]3 years ago
5 0

Answer:

Equilibrium constant for the reaction is 0.0892

Explanation:

This is the reaction of equilibrium

         H₂O      +     Cl₂O     ⇄    2HClO

Eq   0.077M       0.077M           0.023M

Let's make the expression for Kc

Kc = [HClO]² / [Cl₂O] . [H₂O]

Kc= 0.023² / 0.077 . 0.077 = 0.0892

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Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

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Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

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Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

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