At high temps carbon reacts with O2 to produce as follows C(s) + O2(g) <> 2CO(g) when 0.350 mol of O2 and excess carbon were placed in a 5.00-L container and heated the equil. concentration of CO was found to be 0.060 M. What is the equil. constant, Kc, for this reaction?

(1) 0.010
(2) 0.072
(3) 1.2
(4) 0.17
(5) 0.090

(5) 0.090

Explanation:

The initial concentration of O₂ is:

0.350 mol / 5.00 L = 7.00 × 10⁻² mol/L

We can find the concentrations at equilibrium using an ICE chart.

      2 C(s)   +   O₂(g)   ⇄   2 CO(g)

I                  7.00 × 10⁻²         0

C                      -x                  +2x

E             7.00 × 10⁻² - x         2x

Since the concentration at equlibrium of CO was 0.060 M,

2x = 0.060 M

x = 0.030 M

The concentrations at equilibrium are:

[O₂] = 7.00 × 10⁻² - x = 7.00 × 10⁻² - 0.030 = 0.040 M

[CO] = 2x = 2 (0.030) = 0.060 M

The equilibrium constant (Kc) is:

Kc = [CO]²/[O₂]

Kc = (0.060)²/(0.040)

Kc = 0.090

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.