Answer:
Option A) CO(g) + 1/2O2(g) <=> CO2(g).
Explanation:
A background knowledge of reaction rates shows that pressure will only affect gaseous reactant.
Further more, we understood that for pressure to effectively affect gaseous molecules, the total volume of the gaseous reactant must be different from the total volume of the gaseous products.
Now, let us consider the equation given in the question:
A) CO(g) + 1/2O2(g) <=> CO2(g).
B) CaCO3(s) <=> CaO(s) + CO2(g).
C) 2H2(g) + O2(g) <=> 2H2O(l).
D) 2Hg(l) + O2(g) <=> 2HgO(s).
E) CO2(g) + H2(g) <=> CO(g) + H2O(g).
From the above, only option A and E has gaseous reactant and product.
For option A:
CO(g) + 1/2O2(g) <=> CO2(g).
Total volume of reactant = 1 + 1/2 = 3/2 L
Total volume of product = 1 L
Since the volume of the reactant and that of the product are different, therefore, a change in pressure will affect the reaction.
For option E:
CO2(g) + H2(g) <=> CO(g) + H2O(g).
Total volume of reactant = 1 + 1 = 2 L
Total volume of product = 1 + 1 = 2 L
Since the volume of the reactant and that of the product are the same, therefore, a change in pressure will have no effect in the reaction.
The equilibrium equation, CO2(g) + H2(g) ⇄ CO(g) + H2O(g), is not affected by pressure changes at constant temperature because there are equal volumes of reactants and products on both sides of reaction equation.
For a gas phase reaction, changes in pressure would affect the direction in which the reaction moves. When the pressure is increased, the reaction moves in the direction of lesser volumes. When the pressure is decreased, the reaction moves in the direction of greater volumes.
For the reaction; CO2(g) + H2(g) ⇄ CO(g) + H2O(g), there are equal volumes of reactants and products on either side of the reaction equation. Therefore, the equilibria would not be affected by pressure changes at constant temperature.
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