# 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.

Jun 26, 2017

${Q}_{c} = 1.15$

#### 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 $1$ ${\text{mol H}}_{2}$ is added (from the $1$ $\text{L}$-container), there will now be $3.00$ total moles of ${\text{H}}_{2}$.

The concentrations of each species at this point are

• $\text{H"_2"O}$: (2.00color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 0.667M

• ${\text{N}}_{2}$: (2.60color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 0.867M

• ${\text{H}}_{2}$: (3.00color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 1.00M

• $\text{NO}$: (2.00color(white)(l)"mol")/(color(green)(3)color(white)(l)color(green)("L")) = 0.667M

The reaction quotient expression for this reaction is

${Q}_{c} = \left(\left[{\text{H"_2]^2["NO"]^2)/(["H"_2"O"]^2["N}}_{2}\right]\right)$

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