Consider the following equilibrium reaction: \(2AB_{{2}{(g)}} + B_{{2}{(g)}} \to 2AB_{{3}{(g)}}\). \(\Delta H= -X kJmol^{-1}\). The backward reaction will be favored by
-
a decrease in pressure
- an increase in pressure
- a decrease in temperature
- an introduction of a positive catalyst
The correct answer is: A
Explanation
The equilibrium position is shifted to the left with a decrease in pressure in this system because the number of gaseous molecules on the right is less. (Le-Chatelier's principle).