# Question 90689

Aug 7, 2017

Here's what I got.

#### Explanation:

If I understand this correctly, the problem tells you that at a certain temperature, the following equilibrium reaction

${\text{N"_ (2(g)) + color(red)(3)"H"_ (2(g)) rightleftharpoons color(blue)(2)"NH}}_{3 \left(g\right)}$

involves a total of $2$ moles of substance at equilibrium, out of which $0.2$ moles are hydrogen gas and $0.1$ moles are nitrogen gas.

Right from the start, you can use this info to say that the reaction vessel will contain

${\text{2 moles" - ("0.2 moles H"_2 - "0.1 moles N"_2) = "1.7 moles NH}}_{3}$

Now, you know that the volume of the reaction vessel is equal to ${\text{500 cm}}^{3}$, so use this to calculate the equilibrium concentrations of the three chemical species involved in the reaction.

["H"_2] = "0.2 moles"/(500 * 10^(-3)"dm"^3) = "0.4 mol dm"^(-3)

["N"_2] = "0.1 moles"/(500 * 10^(-3)"dm"^3) = "0.2 mol dm"^(-3)

["NH"_3] = "1.7 moles"/(500 * 10^(-3)"dm"^3) = "3.4 mol dm"^(-3)#

By definition, the equilibrium constant for this reaction will take the form

${K}_{c} = \left(\left[{\text{NH"_3]^color(blue)(2))/(["H"_2]^color(red)(3) * ["N}}_{2}\right]\right)$

Plug in your values to find--I'll calculate the value of the equilibrium constant without added units

${K}_{c} = \frac{{3.4}^{\textcolor{b l u e}{2}}}{{0.4}^{\textcolor{red}{3}} \cdot 0.2} = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{900}}}$

The answer is rounded to one significant figure.