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.

1 Answer
Jun 26, 2017

Answer:

#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#