Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−). I2(s) + I−(aq) ⇌ I3−(aq) When 0.0800 moles of iodine crystals and 0.2400 moles of iodide ions were added to deionised water and made up to a litre of solution, 0.0793 moles of triiodide ions (I3−) were present at equilibrium. Write the equilibrium constant (Kc) expression for this equilibrium reaction. Calculate the value of the equilibrium constant (Kc) for the reaction at room temperature. State and explain the effect on the equilibrium concentration of triiodide ions of adding a substance that reacts with iodine, e.g. starch.

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Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−) - Leaving Cert Chemistry - Question b - 2012

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Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−). I2(s) + I−(aq) ⇌ I3−(... show full transcript

Worked Solution & Example Answer:Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−) - Leaving Cert Chemistry - Question b - 2012

Step 1

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

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Answer

The equilibrium constant expression for the reaction is given by:

$K_c = \frac{[I_3^-]}{[I_2][I^-]}$

where [I3−] is the concentration of triiodide ions, [I2] is the concentration of iodine, and [I−] is the concentration of iodide ions.

Step 2

Calculate the value of the equilibrium constant (Kc) for the reaction at room temperature.

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Answer

To calculate the equilibrium concentrations, we must first determine the initial concentrations based on the moles given:

Initial moles of I2 = 0.0800 M in 1 L = 0.0800 M
Initial moles of I− = 0.2400 M in 1 L = 0.2400 M
At equilibrium, moles of I3− = 0.0793 M

Using the change in concentration to find equilibrium concentrations:

For I2, the change is:
- 0.0800 M - x = 0.0800 M - 0.0793 M = 0.0007 M
For I−:
- 0.2400 M - x = 0.2400 M - 0.0793 M = 0.1607 M

Thus the equilibrium concentrations are:

[I2] = 0.0007 M
[I−] = 0.1607 M
[I3−] = 0.0793 M

Now substituting values into the equilibrium constant expression:

$K_c = \frac{0.0793}{0.0007 \times 0.1607} = \frac{704.95}{705}$

So, the equilibrium constant Kc is approximately 704.95.

Step 3

State and explain the effect on the equilibrium concentration of triiodide ions of adding a substance that reacts with iodine, e.g. starch.

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Answer

When a substance that reacts with iodine, such as starch, is added, the equilibrium concentration of triiodide ions (I3−) will decrease.

This occurs because the addition of starch effectively removes some iodine from the solution. According to Le Chatelier's principle, the equilibrium will shift to the left (toward reactants) to restore the concentration of iodine, resulting in a decrease in the concentration of triiodide ions. The system seeks to counteract the change, thus shifting back to re-establish equilibrium.

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Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−) - Leaving Cert Chemistry - Question b - 2012

Worked Solution & Example Answer:Consider the following room temperature equilibrium reaction used to dissolve iodine (I2) crystals in an aqueous solution of iodide ions (I−) - Leaving Cert Chemistry - Question b - 2012

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

Calculate the value of the equilibrium constant (Kc) for the reaction at room temperature.

State and explain the effect on the equilibrium concentration of triiodide ions of adding a substance that reacts with iodine, e.g. starch.

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