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.
