N\(_2\)O\(_4\),(g) ↔ 2NO\(_2\)(g)
In the endothermic reaction above, more product formation will be favoured by
a constant volume
an increase in pressure
a decrease in pressure
a decrease in volume
Explanation
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Discussions (7)
An increase in pressure of a gas equillibrum reaction favours side with lowest number of mole/volume while decrease in pressure favours side with highest number of moles/volume.4rm d equation above,d product has highest number of mole so decrease in pressure favours product formation,option c is correct
C is correct, increase in pressure favors side with lower volume and vice versa.
Guys this is endothermic reaction it's B
Probably they made a typo if it's Exothermic it's C
if there is an decrease in press more product if the forward reaction produce an increase in volume
C is totaly wrong...the reaction above is endothermic as shown in mole representation, but this reaction involves only gases and pressure increase favours forward reaction...if u don't agree then go screw yourself.
