Question

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 0.0280 mol gaseous NH3 and 0.0120 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 3.00×10-3 M. Calculate Keq for the reaction at this temperature.

Answer 1

Answer: The value of
`K_(eq)` is
`4.84* 10^(-5)`

Step-by-step explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 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 =
`(0.0280)/(1.00)=0.0280M`

Concentration of oxygen gas =
`(0.0120)/(1.00)=0.0120M`

The given chemical equation follows:

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

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas =
`3.00* 10^(-3)M=0.003`

Evaluating the value of 'x', we get:

`\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M`

Now, equilibrium concentration of ammonia =
`0.0280-4x=[0.0280-(4* 0.0015)]=0.022M`

Equilibrium concentration of oxygen gas =
`0.0120-3x=[0.0120-(3* 0.0015)]=0.0075M`

Equilibrium concentration of water =
`6x=(6* 0.0015)]=0.009M`

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

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

Putting values in above expression, we get:

`K_(eq)=((0.009)^6* (0.003)^2)/((0.022)^4* (0.0075)^3)\\\\K_(eq)=4.84* 10^(-5)`

Hence, the value of
`K_(eq)` is
`4.84* 10^(-5)`