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QUESTION

# Under certain conditions, nitrogen and oxygen react to form dinitrogen oxide. Suppose the mixture of 0.482 mol N2 and 0.933 mol O2 is placed in a reaction vessel of volume 10.0 L and allowed to form N2O at a temperature for which Kc = 2.0x10^-37?

I assume you are looking for the final equilibrium concentrations of this reaction.

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 .

" " " "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.

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