Question

Changes in pressure can have a large effect on equilibrium systems containing gaseous components.

1. changing the concentration of gaseous components
2. adding an inert gas has no effect since the gas does not take part in the reaction, all partial pressures stay the same
3.changing the volume of the reaction vessel. This will cause a shift in the equilibrium position if the number of moles of gas is different on the reactant and product side (so Δn = n products - n reactants)

How would you change the volume for each of the following reactions to increase the yield of the product(s)?1. CaCO3(s) ⇋ CaO(s) + CO2(g) (increase, decrease, no change)2. S(s) + 3F2(g) ⇋ SF6(g) (increase, decrease, no change)3. Cl2(g) + I2(g) ⇋ 2ICl(g) (increase, decrease, no change)

Answer 1

Answer:

The correct option is 1, since by changing the partial pressures the gas pressures change, the gases go from the zones of higher partial pressure to the zones of lower partial pressure, an example of this is the homeostasis of the human pulmonary alveolus in gas exchange with CO2 and O2.

Explanation:

In the first it increases, in the second the volume is maintained, and in the third reaction it decreases.

Answer 2

Answer:

1. Increase volume.

2. No change.

3. No change.

Explanation:

Hello,

In this case, if we want to shift the reaction rightwards, based on the Le Chatelier's principle we would have to:

1. For this reaction:

$CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$

- Increase the volume or decrease the pressure, since there are more gaseous moles at the products.

2. For this reaction:

$S(s) + 3F_2(g) \rightleftharpoons SF_6(g)$

- Do nothing since it is not possible to achieve it as we have the same number of gaseous moles at both reactants and products.

3. For this reaction:

$Cl_2(g) + I_2(g)\rightleftharpoons 2ICl(g)$

- Do nothing since it is not possible to achieve it as we have the same number of gaseous moles at both reactants and products.

Regards.