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
The correct answer is: C
Explanation
In the endothermic reaction, an increase in pressure will favour the side with the lesser value of number of moles of the gaseous specie while a decrease in pressure will favour the side with a higher value of number of moles of the gaseous specie.
From the above equation, the product(on the RHS) has a higher value of number of moles of NO\(_2\) i.e 2 moles, when compared to the number of moles of N\(_2\)O\(_4\), on the LHS. i.e 1 mole of N\(_2\)O\(_4\), Therefore, a decrease in pressure will favour more product formation.