In: Chemistry
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?
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