# Question #1eb65

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

I'm not sure I followed what you did there, so I'll just solve it completely here.

So, you know that you're dealing with an equilibrium reaction between nitrogen gas,

Notice that the **equilibrium constant ** for this reaction, *smaller* than **left**, i.e. it will favor the reactants more than the product.

Now, your reaction starts with **only** product, which means that you can expect the **reverse reaction**, i.e. the one that leads to the consumption of

The magnitude of the equilibrium constant tells you that **most of the** nitric oxide will be converted to nitrogen and oxygen, which implies that at equilibrium the concentrations of nitrogen and oxygen will be *significantly larger* than that of nitric oxide.

Set up your **ICE table** like this

#" " "N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO"_text((g])#

By definition, the equilibrium constant for this reaction will be

#K_c = (["NO"]^color(red)(2))/( ["N"_2] * ["O"_2]) = (0.500 - color(red)(2)x)^color(red)(2)/(x * x)#

#K_c = (0.500 - color(red)(2)x)^color(red)(2)/x^2#

Now, notice that you can take the square root of both sides to get

#sqrt(K_c) = sqrt((0.500 - color(red)(2)x)^color(red)(2)/x^2)#

#sqrt(1.0 * 10^(-5)) = (0.500 - 2x)/x#

Rearrange this equation to solve for

#sqrt(10^(-5)) * x = 0.500 - 2x#

#x * (2 + sqrt(10^(-5))) = 0.500#

#x = 0.500/(2 + sqrt(10^(-5))) ~~ 0.2499988#

Notice how small the equilibrium concentration of nitric oxide will be when compared with those of nitrogen and oxygen

#["NO"] = 0.500 - 2 * 0.2499988= 2.4 * 10^(-6)"M"#

#["N"_2] = ["O"_2] = x = "0.2499988 M" ~~ "0.25 M"#

Practically, almost all of the nitric oxide will be converted to nitrogen and oxygen.

So remember, if you only start with products, the equilibrium will **automatically** shift to favor the formation of the reactants. Likewise, if you start with only **reactants**, the equilibrium will **automatically** shift to favor the formation of the products.

The *magnitude of the shift* will depend on the value of the **equilibrium constant**,