Question #90689

1 Answer
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

"N"_ (2(g)) + color(red)(3)"H"_ (2(g)) rightleftharpoons color(blue)(2)"NH"_ (3(g))

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

"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 "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 = (["NH"_3]^color(blue)(2))/(["H"_2]^color(red)(3) * ["N"_2])

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

K_c = (3.4^color(blue)(2))/(0.4^color(red)(3) * 0.2) = color(darkgreen)(ul(color(black)(900)))

The answer is rounded to one significant figure.