The termination step of the free-radical chlorination of methane is the most stable one among all three steps.
The free-radical substitution reaction between chlorine and methane features three major steps:
Initiation, during which chlorine molecules undergo homolytic fission to produce chlorine free radicals. Ultraviolet radiations are typically applied to supply the energy required for breaking the chlorine-chlorine single bonds. The initiation step is thus <em>endothermic</em>.
Propagation, a process in which chlorine free radicals react with methane molecules and remove a hydrogen atom from the alkane to produce hydrogen chloride and an alkyl radical e.g.,
. The carbon-containing free radical would react with chlorine molecules to produce chloromethane and yet another chlorine free radical. This process can well repeat itself to chlorinate a significant number of methane molecules.
Termination. Free radicals combine to produce molecules. For example, two chlorine free radicals would combine to produce a chlorine molecule, whereas two alkyl free radicals would combine to produce an alkane with two-carbon atoms in its backbone.
Chemical processes that increase the stability of a substance reduces its chemical potential energy. Energy conserves, thus such processes would also release energy equal to the potential energy lost in quantity. Free radicals are unstable and- as seen in the propagation step- compete readily with neutral molecules for their electrons. The propagation step keeps the number of free radicals constant and is therefore more exothermic than the initiation step. The termination step reduces the number of free radicals, increase the stability of the system by the greatest extent, and is therefore the most exothermic step among the three.
Heat energy released : 167.2 kJ
<h3>Further explanation</h3>
Heat can be calculated using the formula:
Q = mc∆T
Q = heat, J
m = mass, g
c = specific heat, joules / g ° C
∆T = temperature difference, ° C / K
m = 2000 g
c = 4.18 J/ g ° C
∆t = 20 ° C

Answer: 128 g/mol
Explanation:
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of the molar mass of its particles.
Mathematically, that is:

Since, you know the ratio of two rates and the molar mass of one gas, you can calculate the molar mass of the other gas.
The molar mass of the oxygen molecule, O₂ = 2×16.0g/mol = 32.0 g/mol.
In the coming equations, I will use 32 g/mol for simplicity of writing.

So, the molecular mass of the unnknown gas is 128 g/mol.