Consider the equilibrium represented by the following balanced equation. 2BrCl_(g) ⇌ Br2_(g) + Cl2_(g) yellow ↔ red-brown ↔ green (i) Write the equilibrium constant (Kc) expression for this reaction. (ii) Calculate the percentage decomposition into bromine and chlorine of 0.200 moles of BrCl, placed initially in a 5 litre closed container, and allowed to reach equilibrium at 1200 °C according to the equation above. Take the value of Kc at this temperature as 0.220. Give your answer correct to the nearest whole number. (iii) Increasing the pressure on this equilibrium mixture at 1200 °C, intensifies the colour of the mixture but does not change the percentage dissociation of BrCl. Explain.

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Consider the equilibrium represented by the following balanced equation - Leaving Cert Chemistry - Question b - 2019

Question b

Consider the equilibrium represented by the following balanced equation. 2BrCl_(g) ⇌ Br2_(g) + Cl2_(g) yellow ↔ red-brown ↔ green (i) Write the equilibrium consta... show full transcript

Worked Solution & Example Answer:Consider the equilibrium represented by the following balanced equation - Leaving Cert Chemistry - Question b - 2019

Step 1

Write the equilibrium constant (Kc) expression for this reaction.

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Answer

The equilibrium constant expression for the given reaction can be written as:

$K_c = \frac{[Br_2][Cl_2]}{[BrCl]^2}$

Here, ([Br_2]) and ([Cl_2]) are the equilibrium concentrations of bromine and chlorine, respectively, while ([BrCl]) is the concentration of bromine monochloride.

Step 2

Calculate the percentage decomposition into bromine and chlorine.

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Answer

Let ( x ) be the number of moles of BrCl that decomposed.

Initially:

Moles of BrCl = 0.200 moles
Moles of Br2 and Cl2 = 0

Change:

BrCl = -2x
Br2 = +x
Cl2 = +x

Equilibrium:

BrCl = 0.200 - 2x
Br2 = x
Cl2 = x

Now substitute these values into the Kc expression:

$K_c = 0.220 = \frac{x \cdot x}{(0.200 - 2x)^2} = \frac{x^2}{(0.200 - 2x)^2}$

This leads to:

$0.220(0.200 - 2x)^2 = x^2$

Expanding and solving the quadratic will yield:

After calculation, we find that ( x \approx 0.0968 ) moles.

Now, to find the percentage decomposition:

( ext{Percentage decomposed} = \frac{x}{0.200} \times 100 \approx 48.4% \approx 48% \text{ (nearest whole number)})

Step 3

Explain why increasing the pressure on this equilibrium mixture at 1200 °C intensifies the colour of the mixture but does not change the percentage dissociation of BrCl.

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Answer

Increasing the pressure of a gaseous reaction will favor the side with fewer moles of gas due to Le Chatelier's principle. In this reaction, the left side has 2 moles of BrCl while the right side has 3 moles (1 Br2 and 1 Cl2). Therefore, increasing pressure would not favor the forward reaction (for decomposition towards Br2 and Cl2), but it would intensify the existing colors due to concentration changes without affecting the percentage dissociation of BrCl. The position of equilibrium doesn't shift to decompose more BrCl; instead, the existing concentrations of products increase due to the reduced volume available in the container.

Consider the equilibrium represented by the following balanced equation - Leaving Cert Chemistry - Question b - 2019

Worked Solution & Example Answer:Consider the equilibrium represented by the following balanced equation - Leaving Cert Chemistry - Question b - 2019

Write the equilibrium constant (Kc) expression for this reaction.

Calculate the percentage decomposition into bromine and chlorine.

Explain why increasing the pressure on this equilibrium mixture at 1200 °C intensifies the colour of the mixture but does not change the percentage dissociation of BrCl.

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