Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 7.50 l tank with 29.0 mol of ammonia gas at 35.0 °C. She then raises the temperature, and when the mixture has come to equilibrium measures the amount of nitrogen gas to be 13.0 mol.
Calculate the concentration equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. Round your answer to significant digits.

The equilibrium constant for the reaction is 2.47

Explanation:

We are given:

Initial moles of ammonia gas = 29.0 moles

Equilibrium moles of nitrogen gas = 13.0 moles

Volume of the tank = 7.50 L

Molarity is calculated by using the formula:

The chemical equation for the decomposition of ammonia follows:

Initial:             3.87

At eqllm:       3.87-2x    x      3x

Evaluating the value of 'x'

So, equilibrium concentration of ammonia = (3.87 - 2x) = [3.87 - 2(0.577)] = 2.716 M

Equilibrium concentration of nitrogen gas = x = 0.577 M

The expression of for above equation follows:

Putting values in above equation, we get:

Hence, the equilibrium constant for the reaction is 2.47

practical process described pest part of scientific investigation.

because practical knowledge is powerful then therotical knowledge.