The following reaction establishes equilibrium at 2000 K: If the reaction began with 0.110 of and 0.110 of , what were the equilibrium concentrations of all species

[N₂] = 0.098M

[O₂] = 0.098M

[NO] = 0.004M

Explanation:

The reaction is:

N₂(g) + O₂(g) ⇄ 2 NO(g)

Where K of equilibrium is:

4.1x10⁻⁴ = [NO]² / [N₂] [O₂]

Concentrations of each species are concentrations in equilibrium.

If some N₂ and O₂ gases reacts producing NO, the equilibrium concentrations are:

[N₂] = 0.100M - X

[O₂] = 0.100M - X

[NO] = 2X

Only 1 mole of N₂ and O₂ reacts producing 2 moles of NO

Replacing:

4.1x10⁻⁴ = [NO]² / [N₂] [O₂]

4.1x10⁻⁴ = [X]² / [0.100-X] [0.100-X]

4.1x10⁻⁴ = [X]² / [X²-0.2X + 0.01]

4.1x10⁻⁴X² - 8.2x10⁻⁵X + 4.1x10⁻⁶ = [X]²

-0.99959X² - 8.2x10⁻⁵ X+ 4.1x10⁻⁶ = 0

Solving for X:

X = -0.002. False solution. There is no negative concentrations.

X=  0.002. Right solution.

[N₂] = 0.100M - 0.002M = 0.098M

[O₂] = 0.100M - 0.002M = 0.098M

[NO] = 2*0.002 = 0.004M

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a

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.