Question

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Answer 1

Answer: The value of
`K_(eq)` is
`4.66* 10^(-5)`

Step-by-step explanation:

We are given:

Initial moles of ammonia = 0.0120 moles

Initial moles of oxygen gas = 0.0170 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

`\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}`

So, concentration of ammonia =
`(0.0120)/(1.00)=0.0120M`

Concentration of oxygen gas =
`(0.0170)/(1.00)=0.0170M`

The given chemical equation follows:

`4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)`

Initial: 0.0120 0.0170

At eqllm: 0.0120-4x 0.0170-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas =
`2.20* 10^(-3)M=0.00220M`

Evaluating the value of 'x', we get:

`\Rightarrow 2x=0.00220\\\\\Rightarrow x=(0.00220)/(2)=0.00110M`

Now, equilibrium concentration of ammonia =
`0.0120-4x=[0.0120-(4* 0.00110)]=0.00760M`

Equilibrium concentration of oxygen gas =
`0.0170-3x=[0.0170-(3* 0.00110)]=0.0137M`

Equilibrium concentration of water =
`6x=[6* 0.00110]=0.00660M`

The expression of
`K_(eq)` for the above reaction follows:

`K_(eq)=([N_2]^2* [H_2O]^6)/([NH_3]^4* [O_2]^3)`

Putting values in above expression, we get:

`K_(eq)=((0.00220)^2* (0.00660)^6)/((0.00760)^4* (0.0137)^3)\\\\K_(eq)=4.66* 10^(-5)`

Hence, the value of
`K_(eq)` is
`4.66* 10^(-5)`