Question

Each of the following reactions is allowed to reach equilibrium in a sealed container. For which of the reactions could you shift the equilibrium to the right by decreasing the pressure?
CH4(g) + 2O2(g) Double arrow yields CO2(g) + 2H2O(l)

CaCO3(s) Double arrow yields CaO(s) + CO2(g)

Br2(g) + 3Cl2(g) Double arrow yields 2BrCl3(g)

2H2S(g) + 3O2(g) Double arrow yields 2SO2(g) + 2H2O(g)

Answer 1

Answer:

None of the reaction will be favored to the right by a decrease in pressure.

Explanation:

       CH₄(g) + 2O₂(g) ⇄ CO₂(g) + 2H₂O(l)  

       CaCO₃(s) ⇄  CaO(s) + CO₂(g)  

       Br₂(g) + 3Cl₂(g) ⇄  2BrCl₃(g)  

       2H₂S(g) + 3O₂(g) ⇄  2SO₂(g) + 2H₂O(g)

From Le Chatellier's principle, we must understand that pressure changes only affects reactions in gaseous phases. The second reaction will not be affected by pressure.

We are now left with three equations.

Also, increase in pressure favors sides with lower volume. We can know the volume from the coefficients in the equation. Now let us check the volumes:

          CH₄(g) + 2O₂(g) ⇄ CO₂(g) + 2H₂O(l)  

         3 moles of gases      3 moles of gases

        Br₂(g) + 3Cl₂(g) ⇄  2BrCl₃(g)  

         4 moles of gases     2 moles of gases

       2H₂S(g) + 3O₂(g) ⇄  2SO₂(g) + 2H₂O(g)

        5 moles of gases      4 moles of gases

None of the reaction will be favored to the right by a decrease in pressure.

The first reaction will not be affected by any change in pressure because the total number of moles on the two sides are equal.

The last two reactions will be favored to the right by increasing pressure and a decrease in pressure will favor the backward left reaction.

Answer 2

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

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the pressure is decreased, the volume will increase according to Boyle's Law. Now, according to the Le-Chatelier's principle, the equilibrium will shift in the direction where increase in pressure is taking place, i.e in a direction where number of moles of gases is increasing.

So for the give reactions, the equilibrium will shift in right if the moles of gaseous products is more than the moles of gaseous reactants.

a. $CH_4(g)+2O_2(g)\rightleftharpoons CO_2(g)+2H_2O(l)$

The number of moles of gases are equal on both sides.

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

There is 1 mole of gaseous product and 0 moles of gaseous reactants.

c. $Br_2(g)+3Cl_2(g)\rightleftharpoons 2BrCl_3(g)$

There are 2 moles of gaseous products and 4 moles of gaseous reactants.

d. $2H_2S(g)+3O_2(g)\rightleftharpoons 2SO_2(g)+2H_2O(g)$

There are 4 moles of gaseous products and 5 moles of gaseous reactants.