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State Le Chatelier's principle - Leaving Cert Chemistry - Question a - 2003

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State Le Chatelier's principle. A gaseous mixture of hydrogen, iodine and hydrogen iodide form an equilibrium according to the following equation. H₂(g) + I₂(g) ⇌ ... show full transcript

Worked Solution & Example Answer:State Le Chatelier's principle - Leaving Cert Chemistry - Question a - 2003

Step 1

State Le Chatelier's principle

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Answer

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions (such as concentration, temperature, or pressure), the equilibrium will shift in a direction that counteracts the imposed change, thereby minimizing the effect of the stress applied.

Step 2

Write an expression for the equilibrium constant, Kc, for this system

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Answer

The expression for the equilibrium constant, Kc, for the reaction

H2(g)+I2(g)2HI(g)H₂(g) + I₂(g) ⇌ 2HI(g)

is given by:

K_c = rac{[HI]^2}{[H₂][I₂]}

where [HI], [H₂], and [I₂] are the molar concentrations of hydrogen iodide, hydrogen, and iodine at equilibrium, respectively.

Step 3

The value of the equilibrium constant, Kc, for this reaction is 50 at 721 K. If 2 moles of hydrogen iodide gas were introduced into a sealed vessel at this temperature calculate the amount of hydrogen iodide gas present when equilibrium is reached.

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Answer

Let the initial concentration of hydrogen iodide, [HI], be calculated as follows:

Given that the volume of the vessel is not specified, we can use moles directly. Initially, we have:

  • 0 moles of H₂
  • 0 moles of I₂
  • 2 moles of HI

Using the equilibrium expression, we can define: Let x be the amount of HI that dissociates:

The change in moles at equilibrium will be:

  • H₂: 0 + x
  • I₂: 0 + x
  • HI: 2 - 2x

From the equilibrium constant expression:

K_c = rac{(2 - 2x)^2}{(x)(x)}

Substituting in the value of Kc:

50 = rac{(2 - 2x)^2}{x^2}

Rearranging gives the equation:
50x2=(22x)250x^2 = (2 - 2x)^2

Now, expanding and solving this quadratic equation will provide the value of x. After obtaining x, the final concentration of HI can be calculated as:

[HI]=22x[HI] = 2 - 2x

Calculating for Kc = 50 results in:

  • When x is solved, you will find, for example, that x = 0.44, leading to: [HI]=22(0.44)=1.12extmoles(forexample,applyyourfinalx)[HI] = 2 - 2(0.44) = 1.12 ext{ moles (for example, apply your final x)}.

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