For the first-order reaction, 2 n2o(g) → 2 n2(g) + o2(g), what is the concentration of n2o after 3 half-lives if 0.25 mol of n2o is initially placed into a 1.00-l reaction vessel?

The concentration of after 3 half-lives is 0.03125 M.

Explanation:

We are given:

Moles of = 0.25 moles

Volume of solution = 1 L

Putting values in above equation, we get:

where,

a = concentration of reactant left after n-half lives = ?

= Initial concentration of the reactant =  0.25 mol/L

n = number of half lives  = 3

Putting values in above equation, we get:

Hence, the concentration of after 3 half-lives is 0.03125 M.

in chemistry, a solution’s concentration is how much of a dissolvable substance, known as a solute, is mixed with another substance, called the solvent. the standard formula is c = m/v, where c is the concentration, m is the mass of the solute dissolved, and v is the total volume

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.