Question

Consider this reaction: \[{N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g).\] The equilibrium constant ${K_p}$ for the reaction is \[1.0 \times {10^{ - 15}}\] at \[25^\circ C\] and \[0.050\;\] at \[2200^\circ C\]. Is the formation of nitric oxide endothermic or exothermic?

Answer 1

Hint: When a chemical process reaches equilibrium, the equilibrium constant provides information on the relationship between the products and reactants. The equilibrium constant can be defined as the ratio between the amount of reactant and the amount of product used to determine chemical behaviour in a chemical reaction.

Complete answer:
Complete step-by-step answer
\[{N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g).\]
the equilibrium constant \[{K_p}\] is equal to
${K_p} = \dfrac{{{{(NO)}^2}}}{{\left( {{N_2}} \right) \cdot \left( {{O_2}} \right)}}$
Now, at ${25^ \circ }C$ , we have
${K_p} = 1 \times {10^{ - 15}}$
The extremely small value of the equilibrium constant indicates that the equilibrium is substantially to the left, implying that nitrogen and oxygen partial pressures are significantly greater than nitric oxide partial pressures.
In other words, there is much more nitrogen and oxygen gas in the reaction vessel.
At ${2200^ \circ }C$ we know that,
${K_p} = 0.050 = 5.0 \times {10^{ - 2}}$
The equilibrium constant remains 1, indicating that the equilibrium is located to the left. However, you'll observe that as the temperature increases, the value of the equilibrium constant rises as well.
This means that nitric oxide partial pressure increased while nitrogen and oxygen partial pressures decreased, implying that the process used more nitrogen and oxygen while producing more nitric oxide.
You could say that when the temperature rose, the equilibrium moved to the right. As we know, Le Chatelier's Principle governs equilibrium reactions, which states that systems in equilibrium will react to a stress exerted on their position in such a way that the stress is reduced.
As increasing the temperature caused the equilibrium to shift to the right that is it encouraged the production of nitric oxide. In other words, the reaction consumed some of the added heat by producing more nitric oxide.
This implies that you have heat as a reactant and hence the reaction is endothermic
${N_{2(g)}} + {O_{2(g)}} + heat{\text{ }} \rightleftharpoons 2{\text{N}}{{\text{O}}_{(g)}}$ is endothermic reaction.

Note:
Changes in concentration, pressure, temperature, and the presence of inert gases can modify the equilibrium, favouring either forward or backward reaction but not the equilibrium constant.