N\(_2\)O\(_4\) ⇔ 2NO\(_2\) ...
N\(_2\)O\(_4\) ⇔ 2NO\(_2\) (Δ = -ve)
From the reaction above, which of these conditions would produce the highest equilibrium yield for N\(_2\)O\(_4\)?
Low temperature and high pressure
Low temperature and low pressure
high temperature and low pressure
high temperature and high pressure
Correct answer is D
high temperature and high pressure favor the highest yield for N\(_2\)O\(_4\)
high temperature favours the production of the reactant since it is endothermic, while high pressure favours the side involving a decrease in volume.
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