The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually

Question

The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reactions of a number of such steps for methane gas is
CH4(g)+5O2(g)+5NO(g)CO2(g)+H2O(g)+5NO2(g)+2OH(g)
Suppose that an atmospheric chemist combines 145 mL of methane at STP, 885 mL of oxygen at STP, and 59.5 mL of NO at STP in a 1.8 L flask. The reaction is allowed to stand for several weeks at 275 K .

If the reaction reaches 92.0% of completion (92.0% of the limiting reactant is consumed), what are the partial pressures of each of the reactants in the flask at 275 K ?

If the reaction reaches 92.0% of completion (92.0% of the limiting reactant is consumed), what are the partial pressures of each of the products in the flask at 275 K ?

What is the total pressure in the flask?

Explanation / Answer

The gases are at the same T and P since they are in the same container.
Their volumes will react in the same proportions as the coefficients in the balanced equation.
So 145mL CH4 will require 145mL *5 =725 of O2 and 725mL of NO

But only 59.5 mL NO is present and it will be completely consumed. NO is the limiting reactant.

So 0.92 x 59.5 mL = 54.74mL NO reacted.

Amount of unreacted NO= 59.5-54.74 = 4.76 mL

54.74 mL NO reacts with 54.74mL of O2.

Amount of unreacted O2 = 885 -54.74 mL= 830.26 mL

54.74 mL NO reacts with 54.74/5 = 10.95 mL CH4. Amount of unreacted CH4 = 145-10.95 = 134.05 mL
The products will form according to the proportion in the balanced equation.

Amount of CO2 = 54.74 mL

Amount of H2O = 54.74 mL

Amount of NO2 = 54.74 * 5 = 273.7 mL

Amount of OH(g) = 109.48 mL

Total volume of the unreacted gases + products = 1461.73 mL =1.462 L

moles of gas present = 1.462/22.4 = 0.0653 moles (1mole of gas at STP occupies 222.4 L volume)

total pressure of the gas mixture can be calculated by using ideal gas equation.

P = nRT/ V

where, V = 1.462 L, R = 0.082 L. atm/mole/K T= 275 K, n= 0.0653 moles, P= ?

P = 0.0653 moles * 0.082 L. atm/mole/K* 275 K/1.462 L = 1.007 atm

The partial pressures will be proportional to the ratio of the volume of the gas to total volume x total pressure, (since V = nRT/P and RTand P are constants (STP), so n is proportional to V)
P(O2 ) = 830.26 * 1.007/1462 =

P(NO) = 4.76 *1.007/1462 = 0.0033 atm

P(CH4) = 134.05*1.007/1462= 0.092 atm

P(CO2) = 54.74*1.007/1462 = 0.038 atm

P(H2O) = 54.74 *1.007/1462 = 0.038 atm

P(NO2) = 273.7*1.007/1462 = 0.189 atm

P(OH) = 109.48 *1.007/1462 = 0.075 atm

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