How to calculate #K_c#? #"N"_2(g) + "O"_2(g) rightleftharpoons 2"NO"(g) #
The equilibrium concentrations of the gases at 1500K are
O2 = 1.7×10-3M;
N2 = 6.4×10-3M;
NO = 1.1 10-5M.
Calculate the value of Kc at 1500K from these data.
The equilibrium concentrations of the gases at 1500K are
O2 = 1.7×10-3M;
N2 = 6.4×10-3M;
NO = 1.1 10-5M.
Calculate the value of Kc at 1500K from these data.
1 Answer
Explanation:
The equilibrium constant is simply a measure of the position of the equilibrium in terms of the concentration of the products and of the reactants in a given equilibrium reaction.
In other words, the equilibrium constant tells you if you should expect the reaction to favor the products or the reactants at a given temperature.
This is done by comparing the equilibrium concentrations and the stoichiometric coefficients of the chemical species involved in the reaction.
For your equilibrium reaction
#"N"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(red)(2)"NO"_ ((g))#
the equilibrium constant takes the form
#K_c = (["NO"]^color(red)(2))/(["N"_2] * ["O"_2])#
Now, you know that
#["NO"] = 1.1 * 10^(-5)# #"M"# #["N"_2] = 6.4 * 10^(-3)# #"M"# #["O"_2] = 1.7 * 10^(-3)# #"M"#
Right from the start, the fact that, at equilibrium, you have significantly less nitric oxide than nitrogen gas and oxygen gas tells you that at
This means that at this temperature, if you start with nitrogen gas and oxygen gas, only a very small fractions of the molecules will react to form nitric oxide.
Similarly, if you start with nitric oxide, a very large fraction of the molecules to react and form nitrogen gas and oxygen gas.
This means that you should expect to find
#K_c < 1#
which would be consistent with the fact that the equilibrium lies to the left at this temperature.
So, plug in your values to find--I'll skip the added units
#K_c = (1.1 * 10^(-5))^color(red)(2)/(6.4 * 10^(-3) * 1.7 * 10^(-3))#
#K_c = color(darkgreen)(ul(color(black)(1.1 * 10^(-5))))#
As predicted, you have
#K_c < 1#
which confirms why you have higher concentrations of nitrogen gas and oxygen gas than nitric oxide at equilibrium.