# Question edf1e

Aug 24, 2015

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

#### Explanation:

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

$2 {N}_{2} + {O}_{2} r i g h t \le f t h a r p \infty n s 2 {N}_{2} O$

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:

${N}_{2} = \text{0.482 Mole}$, ${O}_{2} = \text{0.933 Mole}$ ${K}_{c} = 2 \times {10}^{- 37}$

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.

$\text{ " " "2N_2" " " " + " " " "O_2" " rightleftharpoons " " " } 2 {N}_{2} O$

$\text{I" " " " " " " "0.482" " " " " " " " " "0.933" " " " " " " " " " } 0$
$\text{C" " " " " " " "(-x)" " " " " " " " " "( -0.5x)" " " " " " " " } \left(+ x\right)$
$\text{E" " " " " (0.482-x)" " "(0.933-0.5x)" " " " " " } \left(+ x\right)$

(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:

$2.0 \times {10}^{-} \left(37\right) = \frac{{\left[{N}_{2} O\right]}^{2}}{{\left[{N}_{2}\right]}^{2} \cdot \left[{O}_{2}\right]}$

Substituting the problem values from the ICE chart:

$2.0 \times {10}^{- 37} = \frac{{x}^{2}}{{\left(0.482 - x\right)}^{2} \cdot \left(0.933 - 0.5 x\right)}$

Solve for $x$ to calculate the final concentrations of ${N}_{2}$, ${O}_{2}$, and ${N}_{2} O$.

Sep 22, 2015