# Use this equation to find the equilibrium constant?

May 8, 2017

Well, the hard part was to come up with the equation for the reaction

$2 \text{N"_2"O"(g) + "O"_2(g) rightleftharpoons 4"NO} \left(g\right)$...

now use it. It is just plugging numbers in and using a calculator:

K_c = (5.6 xx 10^(-9) "M")^4/(("0.0035 M")^2("0.0027 M")) = ???

From where did the exponents originate? On a further note, it is a must to understand what the equation means...

In that regard, what would happen if the concentration of $\text{N"_2"O}$ was doubled? Could ${K}_{c}$ change, or would the equilibrium be disturbed, and then be inclined to shift back to equilibrium and establish the same ${K}_{c}$ again? Which direction would the equilibrium shift if this were to occur (right or left)?