Suppose a 250 mL flask is filled with 0.30 mol of NO and 1.3 mol of NO2. The following reaction becomes possible: $NO_3(g) + NO(g) \rightleftharpoons 2NO_2(g)$. The equilibrium c... Suppose a 250 mL flask is filled with 0.30 mol of NO and 1.3 mol of NO2. The following reaction becomes possible: $NO_3(g) + NO(g) \rightleftharpoons 2NO_2(g)$. The equilibrium constant K for this reaction is 6.29 at the temperature of the flask. Calculate the equilibrium molarity of NO2. Round your answer to two decimal places. Read more

Steps to Solve

1. Calculate initial molarities

To find molarity, we divide the number of moles of each substance by the volume of the flask in liters.

$ [NO]_0 = \frac{0.30 \text{ mol}}{0.250 \text{ L}} = 1.2 \text{ M} $

$ [NO_2]_0 = \frac{1.3 \text{ mol}}{0.250 \text{ L}} = 5.2 \text{ M} $

2. Set up the ICE table

Here's the ICE table for the reaction:

3. Write the equilibrium constant expression

$ K = \frac{[NO_2]^2}{[NO_3][NO]} $

$ 6.29 = \frac{(5.2-2x)^2}{x(1.2+x)} $

4. Solve for x

Expanding and rearranging the equation:

$ 6.29x(1.2+x) = (5.2-2x)^2 $

$ 7.548x + 6.29x^2 = 27.04 - 20.8x + 4x^2 $

$ 2.29x^2 + 28.348x - 27.04 = 0 $

Now, we use the quadratic formula to solve for $x$:

$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

$ x = \frac{-28.348 \pm \sqrt{(28.348)^2 - 4(2.29)(-27.04)}}{2(2.29)} $

$ x = \frac{-28.348 \pm \sqrt{803.6055 + 248.1472}}{4.58} $

$ x = \frac{-28.348 \pm \sqrt{1051.7527}}{4.58} $

$ x = \frac{-28.348 \pm 32.4307}{4.58} $

We have two possible values for $x$:

$ x_1 = \frac{-28.348 + 32.4307}{4.58} = \frac{4.0827}{4.58} = 0.8914 $ $ x_2 = \frac{-28.348 - 32.4307}{4.58} = \frac{-60.7787}{4.58} = -13.2704 $

Since $x$ represents a concentration change, it cannot be negative. Therefore, we take the positive value:

$ x = 0.8914 $

5. Calculate the equilibrium molarity of $NO_2$

$ [NO_2] = 5.2 - 2x = 5.2 - 2(0.8914) = 5.2 - 1.7828 = 3.4172 \text{ M} $

6. Round to two decimal places

$ [NO_2] = 3.42 \text{ M} $