Question

nitrogen and oxygen react to form nitrogen monoxide, like this:

$$ mathrm { n } _ { 2 } ( g ) + mathrm { o } _ { 2 } ( g ) ightarrow 2 mathrm { no } ( g ) $$

the reaction is endothermic. suppose a mixture of $$ mathrm { n } _ { 2 } , mathrm { o } _ { 2 } $$ and no has come to equilibrium in a closed reaction vessel. predict what change, if any, the perturbations in the table below will cause in the composition of the mixture in the vessel. also decide whether the equilibrium shifts to the right or left.

Explanation:

Step1: Consider Le - Chatelier's principle for temperature change in endothermic reactions

For an endothermic reaction \(N_{2}(g)+O_{2}(g)
ightleftharpoons2NO(g)\), heat is absorbed in the forward reaction. When the temperature is raised, according to Le - Chatelier's principle, the system will try to absorb the extra heat. So, the reaction will shift in the forward (right) direction. As the reaction shifts to the right, more \(NO\) is formed. Since pressure of a gas in a mixture (at constant volume) is proportional to its amount (\(P = \frac{nRT}{V}\), \(R\), \(T\), \(V\) constant), the pressure of \(NO\) will increase.

Step2: Consider Le - Chatelier's principle for temperature decrease in endothermic reactions

When the temperature is lowered, the system will try to release heat to counteract the change. For an endothermic reaction, the reverse reaction is exothermic. So, the reaction will shift to the left. As the reaction shifts to the left, \(O_{2}\) (reactant) is formed. Using \(P=\frac{nRT}{V}\) (\(R\), \(T\) (lowered but constant for this analysis), \(V\) constant), the pressure of \(O_{2}\) will increase.

Answer:

perturbation	change in composition	shift in equilibrium
The temperature is lowered	increase	to the left