Question #edf1e

First, you must balance the chemical equation in order to get the coefficients needed for the rate reaction:

#2N_2 + O_2 rightleftharpoons 2N_2O#

Then you need to set up an ICE (Initial, Change, Equilibrium) chart to find the factors that need to be solved in the equation for a numerical result:

#N_2 = "0.482 Mole"#, #O_2 = "0.933 Mole"# #K_c = 2 xx 10^(-37)#

The volume is irrelevant as long as the molar quantities are known and it is a gas mixture. It is only necessary if you are given solution volumes and need to correct from the standard moles/Liter to actual moles in solution for liquid solutions.

#" " " "2N_2" " " " + " " " "O_2" " rightleftharpoons " " " "2N_2O#

#"I" " " " " " " "0.482" " " " " " " " " "0.933" " " " " " " " " " "0#
#"C" " " " " " " "(-x)" " " " " " " " " "( -0.5x)" " " " " " " " "(+x)#
#"E" " " " " (0.482-x)" " "(0.933-0.5x)" " " " " " "(+x)#

(this is why you need to balance the equation, not all of the #O_2# will be used)

The equilibrium equation is

#K_c = (["Products"]^A)/(["Reactants"]^B)" "#, where

#A# and #B# are the coefficients in the balanced chemical equation.

The equation for this system at equilibrium is thus:

#2.0 xx 10^-(37) = ([N_2O]^2)/([N_2]^2 * [O_2])#

Substituting the problem values from the ICE chart:

#2.0 xx 10^(-37) = (x^2)/((0.482-x)^2 * (0.933-0.5x))#

Solve for #x# to calculate the final concentrations of #N_2#, #O_2#, and #N_2O#.

In general, the steps to determine the composition of an equilibrium mixture, given the chemical reaction equation, starting concentrations of the reactants, and the equilibrium constant are the same.

1. Write out the balanced chemical reaction.
2. Set up the I.C.E table to determine the equilibrium concentrations of each species.
3. Use the values from the Equilibrium line of the ICE (including the unknown factors) to set up the equilibrium equation with the equilibrium constant.
4. Solve the equation for the unknown quantity, then back-calculate any other species that may depend on that value (if necessary).