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.

The value of is

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:

So, concentration of ammonia =

Concentration of oxygen gas =

The given chemical equation follows:

Initial:          0.0120    0.0170

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

We are given:

Equilibrium concentration of nitrogen gas =

Evaluating the value of 'x', we get:

Now, equilibrium concentration of ammonia =

Equilibrium concentration of oxygen gas =

Equilibrium concentration of water =

The expression of for the above reaction follows:

Putting values in above expression, we get:

Hence, the value of is

14 days is the half-life of phosphorus-32