Question

Suppose a 250 ml flask is filled with 1.3mol of O2 and 1.5 mol of NO. The following reaction becomes possible: The equilibrium constant for this reaction is at the temperature of the flask. Calculate the equilibrium molarity of . Round your answer to two decimal places.

Answer 1

Answer: The equilibrium molarity of oxygen gas is 7.1 M

Explanation:

We are given:

Moles of oxygen gas = 1.3 mole

Moles of NO = 1.5 moles

Volume of the flask = 250 mL = 0.250 L   (Conversion factor:  1 L = 1000 mL)

To calculate the molarity, we use the equation:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume of solution (in L)}}$

Molarity of oxygen gas = $\frac{1.3}{0.25}=5.2M$

Molarity of NO = $\frac{1.5}{0.25}=6.0M$

The chemical equation follows:

                              $O_2+N_2\rightleftharpoons 2NO$

Initial:                  5.2    -           6.0

At eqllm:          5.2+x     +x      6.0-2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO]}{[O_2][N_2]}$

$K_c=0.394$   (Assuming)

Putting values in above equation, we get:

$0.394=\frac{(6.0-2x)^2}{(5.2+x)\times x}\\\\-3.606x^2+26.0488x-36=0\\\\x=1.9,5.4$

Neglecting the value of x = 5.4 because this cannot be greater than the initial value.

Concentration of oxygen gas at equilibrium = (5.2 + x) = 5.2 + 1.9 = 7.1 M

Hence, the equilibrium molarity of oxygen gas is 7.1 M