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
a decrease in pressure
b
an increase in pressure
c
a decrease in temperature
d
an introduction of a positive catalyst
Explanation
Correct Option
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Siko
1 year ago
increase in pressure will favour the side with the lesser moles which is forward reaction so a decrease in pressure will favour the side with mole number of moles thus the backward reaction hence option A is the correct answer. 
