Question

2 moles of H2O(g), 2.6 moles of N2(g), 2.00 moles of H2(g), and 2.00 moles of NO(g) are in equilibrium in a 2 L container. If a second 1L container with 1 mole of H2(g) is combined with the first, what is the initial reaction quotient?

2 H2O(g) + N2(g) <-----> 2 H2(g) + 2NO(g)

Here is the reaction but I'm completely lost.

Answer 1

`Q_c = 1.15`

Explanation:

We know that the first container is `2` liters in volume, and the one filled with hydrogen is `1` liter, so the combined volume is `color(green)(3` `color(green)("L"`.

And after the `1` `"mol H"_2` is added (from the `1` `"L"`-container), there will now be `3.00` total moles of `"H"_2`.

The concentrations of each species at this point are

• `"H"_2"O"`: `(2.00color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 0.667M`

• `"N"_2`: `(2.60color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 0.867M`

• `"H"_2`: `(3.00color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 1.00M`

• `"NO"`: `(2.00color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 0.667M`

The reaction quotient expression for this reaction is

`Q_c = (["H"_2]^2["NO"]^2)/(["H"_2"O"]^2["N"_2])`

Plugging in the above concentrations, our reaction quotient is

`Q_c = ((1.00M)^2(0.667M)^2)/((0.667M)^2(0.867M)) = color(blue)(1.15`