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
Explanation
Video Explanation
Post your Contribution
Discussions (52)
Them suppose wipe this guy wey the explain this question cord for neck first 
. The correct answer is D.
The correct answer is option D
increase in temperature in an exothermic rxn will favour production of the reactants and high pressure will favour the one with lower moles
The conditions that would produce the highest equilibrium yield for NβOβ are A. Low temperature and high pressure.
The reaction NβOβ β 2NOβ is an exothermic reaction (ΞH = -ve). According to Le Chatelier's principle, a decrease in temperature will shift the equilibrium to the right, favoring the formation of more NβOβ.
Additionally, the reaction involves a decrease in the number of moles of gas (2 moles of NOβ to 1 mole of NβOβ). According to Le Chatelier's principle, an increase in pressure will shift the equilibrium to the side with fewer moles of gas, again favoring 1 the formation of NβOβ.
the correct answer is D, the forward reaction is an exothermic reaction and decrease in temperature favours the exothermic side..we need the high yield of the reactant since it was asked .so it should be High temperature since the reverse is endothermic..and high pressure faviurs the one with lower mole and the mole is 1:2
In an exothermic reaction, an increase in temperature favours the backward reaction leading to the production of high yield of N2O4.
The ratio of volume respectively are 1:2, so a high pressure will make the equillibrium constant shift from right to left.
The answer is high temperature and high pressure(D)
NOTE: the forward reaction in an exothermic system is favoured by lowering the Temperature
Hence this is a reverse case which involves increasing temperature
High temp β favors products (NOβ)
(because the forward reaction releases heat)
Low temp β favors reactant (NβOβ)
(because the system wants to make more heat)
Pressure:
High pressure β favors reactant (NβOβ)
(because it has fewer gas particles)
Low pressure β favors product (NOβ)
(because it has more gas particles)
Correct answer is option A
WHY IS THE VIDEO LESSON SAYING OPTION A IS THE ANSWER AND THE CLASSROOM IS SAYING OPTION D IS THE ANSWER????
IF THE ANSWERS ARE CONTRACDICTING WHY NOT REMOVE THE VIDEO TO AVOID CONFUSION!!!
PS:THE CORRECT ANSWER IS OPTION D
In this case, an increase in temperature will cause the equilibrium position to shift to the left that is it favours reactant formation because it is an exothermic reaction. Again high pressure favours reactant formation because compound with lesser number of gasous concentration is in the reactant side. Option D is the correct π― option.
