Question #1eb65

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Question #1eb65

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Answer 1 · 2015-11-10T03:23:46

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, $N_{2}$ $"N"_2$ , oxygen gas, $O_{2}$ $"O"_2$ , and nitric oxide, $NO$ $"NO"$ .

Notice that the equilibrium constant ** for this reaction, $K_{c}$ $K_c$ , is smaller than $1$ $1$ , which means that the equilibrium will lie to the 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 $NO$ $"NO"$ and the formation of $N_{2}$ $"N"_2$ and $O_{2}$ $"O"_2$ , to be favored.

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

$N_{2(g]} + O_{2(g]} ⇌ 2 {NO}_{(g]}$ $" " "N"_text(2(g]) " "+" " "O"_text(2(g]) " "rightleftharpoons" " color(red)(2)"NO"_text((g])$

$I 00 0.500$ $color(purple)("I")" " " " "0" " " " " " " " " " " "0" " " " " " " " " " 0.500$
$C (+ x) (+ x) (0.500 - 2 x)$ $color(purple)("C")" "(+x)" " " " " "(+x)" " " " (0.500 - color(red)(2)x)$
$E (x) (x) 0.500 - 2 x$ $color(purple)("E")" " " "(x)" " " " " " " "(x)" " " " " "0.500 - color(red)(2)x$

By definition, the equilibrium constant for this reaction will be

$K_{c} = \frac{{[NO]}^{2}}{[N_{2}] \cdot [O_{2}]} = \frac{{(0.500 - 2 x)}^{2}}{x \cdot x}$ $K_c = (["NO"]^color(red)(2))/( ["N"_2] * ["O"_2]) = (0.500 - color(red)(2)x)^color(red)(2)/(x * x)$

$K_{c} = \frac{{(0.500 - 2 x)}^{2}}{x^{2}}$ $K_c = (0.500 - color(red)(2)x)^color(red)(2)/x^2$

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

$\sqrt{K_{c}} = \sqrt{\frac{{(0.500 - 2 x)}^{2}}{x^{2}}}$ $sqrt(K_c) = sqrt((0.500 - color(red)(2)x)^color(red)(2)/x^2)$

$\sqrt{1.0 \cdot 10^{- 5}} = \frac{0.500 - 2 x}{x}$ $sqrt(1.0 * 10^(-5)) = (0.500 - 2x)/x$

Rearrange this equation to solve for $x$ $x$

$\sqrt{10^{- 5}} \cdot x = 0.500 - 2 x$ $sqrt(10^(-5)) * x = 0.500 - 2x$

$x \cdot (2 + \sqrt{10^{- 5}}) = 0.500$ $x * (2 + sqrt(10^(-5))) = 0.500$

$x = \frac{0.500}{2 + \sqrt{10^{- 5}}} \approx 0.2499988$ $x = 0.500/(2 + sqrt(10^(-5))) ~~ 0.2499988$

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

$[NO] = 0.500 - 2 \cdot 0.2499988 = 2.4 \cdot 10^{- 6} M$ $["NO"] = 0.500 - 2 * 0.2499988= 2.4 * 10^(-6)"M"$

$[N_{2}] = [O_{2}] = x = 0.2499988 M \approx 0.25 M$ $["N"_2] = ["O"_2] = x = "0.2499988 M" ~~ "0.25 M"$

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, $K_{c}$ $K_c$ .

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Question #1eb65

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