Question

Consider this reaction: `"N"_ 2(g) + "O"_ 2(g) rightleftharpoons 2"NO"(g)`. The equilibrium constant `K_P`, for the reaction is `1.0 xx 10^-15` at 25°C and `0.050` at 2200°C. Is the formation of nitric oxide endothermic or exothermic?

Answer 1

Endothermic.

Explanation:

The key here is to notice what happens to the equilibrium constant as temperature increases.

You know that for

`"N"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"NO"_((g))`

the equilibrium constant `K_p` is equal to

`K_p = (("NO")^2)/(("N"_2) * ("O"_2))`

`color(red)(!)` Keep in mind that the expression for `K_p` uses the partial pressures of the three chemical species at equilbrium.

Now, at `25^@"C"`, you have

`K_("p 25"^@"C") = 1.0 xx 10^(-15)`

The very, very small value of the equilibrium constant tells you that the equilibrium lies heavily to the left, meaning that the partial pressure of nitrogen gas and of oxygen gas are significantly higher than the partial pressure of nitric oxide.

In other words, the reaction vessel contains significantly more nitrogen gas and oxygen gas.

At `2200^@"C"`, you know that

`K_("p 2200"^@"C") = 0.050 = 5.0 * 10^(-2)`

The equilibrium constant is still `<1`, which means that the equilibrium lies to the left. However, notice that the value of the equilibrium constant increased significantly with an increase in temperature.

This implies that the partial pressure of nitric oxide increased and the partial pressures of nitrogen gas and oxygen gas decreased, i.e. the reaction consumed more nitrogen gas and oxygen gas and produced more nitric oxide.

`{ (("NO")_ (2200^@"C") > ("NO")_ (25^@"C")), (("N"_ 2)_ (2200^@"C") < ("N"_ 2)_ (25^@"C")), (("O"_ 2)_ (220^@"C") < ("O"_ 2)_ (25^@"C")) :} implies color(red)(ul(color(black)(K_ ("p 2200"^@"C") > K_ ("p 25"^@"C"))))`

You can say that an increase in temperature caused to equilibrium to shift to the right.

As you know, equilibrium reactions are governed by Le Chatelier's Principle, which states that systems at equilibrium will react to a stress placed on the position of the equilibrium in such a way as to reduce that stress.

In your case, increasing the temperature caused the equilibrium to shift to the right, i.e. 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, i.e. the reaction is endothermic

`"N"_ (2(g)) + "O"_ (2(g)) + color(red)("heat") rightleftharpoons 2"NO"_((g)) ->` endothermic reaction