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

Correct answer is A

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).

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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

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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