Question

Question #90689

Answer 1

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.