Question

For a hypothetical reaction of the kind
$A{B}_2\left(g\right)+\frac{1}{2}{B}_2\left(g\right)\rightleftharpoons A{B}_3\left(g\right);\Delta H=-xkJ$
More $A{B}_3$ could be produced at equilibrium by:

• A. Using a catalyst
• B. Removing some of $B_2$
• C. Increasing the temperature
• D. Increasing the pressure

Answer 1

The correct option is D Increasing the pressure

We are given an exothermic reaction,
$A{B}_2\left(g\right)+\frac{1}{2}{B}_2\left(g\right)\rightleftharpoons A{B}_3\left(g\right)$

a) Using a catalyst will have no impact on the equilibrium, so cannot increase the amount of $A{B}_3$ using a catalyst.

b) If we remove $B_2$, according to Le Chatelier’s principle the reaction will move in the backward direction, so $A{B}_3$ will further decrease.

c) Since the reaction is exothermic, increasing the temperature will decrease the amount of $A{B}_3$ again.

d) But the number of the gaseous molecules is in the products is lower than the reactants. Increasing the pressure will shift the equilibrium towards the forward reaction and thus will increase the amount of $A{B}_3$.