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

Step-by-step 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 =
`(1.3)/(0.25)=5.2M`

Molarity of NO =
`(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=([NO])/([O_2][N_2])`

`K_c=0.394` (Assuming)

Putting values in above equation, we get:

`0.394=((6.0-2x)^2)/((5.2+x)* 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