Calculate the half life for O* (g) at 25degree C in seconds if the partial press

Question

Calculate the half life for O* (g) at 25degree C in seconds if the partial pressure of water is (0.013 atm). Assume a pseudo-first order rate expression is valid. You do not need any other information to solve this. If the present atmospheric concentration of methane is about 1.8 ppmv and is rising slowly. Emissions of methane into the atmosphere (i.e. the rates) from the most important sources are given in the table below: Using this data, calculate the half-life (t_1/2) of methane in years. Recall that the mass of the atmosphere you calculated early in the semester was 5.3times10^15 tonnes. Note: this is a "back of the envelope" calculation, so approximations are valid. Please identify what assumptions you are using in the solution. The formation of stratospheric ozone has the following rate constant: 0+O_2+MRIGHTARROW O_3 + M k = 1.1times10^-33 cm^6 molec^-1 s^-1 at 220 K What is the rate of this reaction in molec s^-1 if p(total) = 0.010 atm and [O]= 2.1times10^-4 ppm?.

Explanation / Answer

Mass of atm = 5.3 x 1015 ton

= 5.3 x 1021 g

Molar mass of atm = 28.97 g / mol

Moles of atm = 5.3 x 1021 / 28.97 = 1.83 x 1020 mol

Concentration of methane = 1.8 ppmv

Therefore, by Avogadro’s law, moles of methane = 1.8 x 10-6 x 1.83 x 1020 = 3.3 x 1014 mol

Molar mass of methane = 16 g/mol

Mass of methane = 16 x 3.3 x 1014 = 5.28 x 1015 g

Since the concentration is rising very slowly, terfhe influx of methane from various sources (150 + 140 +95 + 45 +170 = 500 million ton / year) should be balanced by an equal rate of decay.

Therefore mass of decayed methane in a year = 500 million ton = 5 x 1014 g

Mass left = 5.28 x 1015 - 5 x 1014 = 4.78 x 1015 g

Fraction remaining = 4.78 / 5.28 = 0.905

Let n be no of half-lives.

n = log 0.905 / log 0.5 = 0.144

1 yr contains 0.144 half-lives

Therefore half-life = 1 / 0.144 = 6.94 yrs

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