NO(g) + CO(g) ↔ (1)/(2)N2(g) + CO2(g) ∆H = 89.3 kj.
What conditions would favour maximum conversion of nitrogen (II) oxide and carbon (II) oxide in the reaction above?
Low temperature and high pressure
High temperature and low pressure
High temperature and high pressure
Low temperature and low pressure
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (21)
C is the correct answer 
you guys should the question properly
Dont get confused of the grammer
The question was how do we obtain more product ,which is N2&CO2
Its an Endothermic reaction so increased in temperature favours foward reaction
The product has the 1.5moles and reactant have 2moles the product has a lower mole,so increased in pressure favours the forward reaction
what are you people saying here abeg, in exothermic reaction, high temperature favors the reactant, then low pressure also favora the reactant because it has more amount of moles abi?
abeg poo
the correct answer should be C.
the reaction is Endothermic, and we are trying to maximise the conversion of NO and CO to N2 and CO2, that is the forward reaction, therefore, increase in temperature favors the forward reaction due to the reaction being endothermic,
¡Hola! Let's analyze the reaction:
NO(g) + CO(g) ⇌ (1/2)N2(g) + CO2(g) ΔH = 89.3 kJ
To favor maximum conversion of nitrogen(II) oxide (NO) and carbon(II) oxide (CO), we need to consider the factors that influence chemical equilibrium:
1. Temperature: Since ΔH is positive (+89.3 kJ), the reaction is endothermic, meaning it absorbs heat. High temperature will favor the forward reaction, increasing the conversion of NO and CO.
2. Pressure: The reaction involves a decrease in the number of moles of gas (2 moles → 1.5 moles), so high pressure will favor the forward reaction, shifting the equilibrium towards the products.
Considering these factors, the conditions that favor maximum conversion of nitrogen(II) oxide and carbon(II) oxide are:
C. High temperature and high pressure
Here's why the other options are not correct:
A. Low temperature and high pressure: Low temperature will slow down the reaction, and high pressure will have a limited effect.
B. High temperature and low pressure: High temperature will favor the reaction, but low pressure will reduce the effect.
D. Low temperature and low pressure: Both conditions will hinder the reaction.
By applying Le Chatelier's principle, we can see that high temperature and high pressure will drive the reaction towards maximum conversion of NO and CO!
increasing the pressure will shift it to d. right favouring forward reaction likewise decreasing the pressure to the left
but for maximum conversion the pressure has to be increased so as not to decrease the reactant on d left
The given reaction is:
NO(g) + CO(g) ⇌ (1/2)N₂(g) + CO₂(g) ΔH = +89.3 kJ (endothermic)
To find the conditions that favor maximum conversion of reactants (NO and CO) to products, consider Le Chatelier’s Principle:
1. Temperature
Since the reaction is endothermic (ΔH is positive), increasing temperature shifts the equilibrium to the right (towards the products).
→ High temperature favors product formation.
2. Pressure
Count the moles of gas:
Reactants: 1 mol NO + 1 mol CO = 2 moles
Products: ½ mol N₂ + 1 mol CO₂ = 1.5 moles
So, the reaction results in fewer gas molecules. According to Le Chatelier’s Principle, increasing pressure favors the side with fewer moles.
→ High pressure favors product formation.
Correct Answer: C. High temperature and high pressure
The answer is C, high temperature favors forward reaction in endothermic reactions. We want the reactacnts to be converted no?
de corret answer to dis is c..bcos dat reaction is exothermic so increase in tempeature favours de backward reaction and increace in pressure favours de one with low concentration of masss
d corect option should b D since d volum of d product is higher dan tat of d reactant
this is because decrease in temperature favours backward reaction but increase in temperature favours forward reaction. While d higher d pressure, d higher d rate of reaction
