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

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.