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 as follows: CH4(g)+5O2(g)+5NO(g)CO2(g)+H2O(g)+5NO2(g)+2OH(g) Suppose that an atmospheric chemist combines 160 mL of methane at STP, 890 mL of oxygen at STP, and 57.0 mL of NO at STP in a 2.0 L flask. The reaction is allowed to stand for several weeks at 275 K.

Part A

If the reaction reaches 90.0% of completion (90.0% of the limiting reactant is consumed), what are the partial pressures of each of the reactants in the flask at 275 K?

Part B

If the reaction reaches 90.0% of completion (90.0% of the limiting reactant is consumed), what are the partial pressures of each of the products in the flask at 275 K?

Part C

What is the total pressure in the flask?

Explanation / Answer

Assuming Ideal gas behaviour,

Ideal gas equation is , P V = n R T

At STP, (Initial)

nCH4 = (1*0.160) / (0.0821*273) = 0.00714 mol

nO2 = (1*0.890) / (0.0821*273) = 0.0397 mol

nNO = (1*0.057) / (0.0821*273) = 0.00254 mol

According to the given stoichiometric equation,

1 mol CH4 = 5 mol O2 = 5 mol NO

So, for 0.00714 mol CH4 = 0.0357 mol O2 = 0.0357 mol NO are required.

Hence the limiting reagent is NO.

90% of limiting reagent is, 0.00254*90/100 = 0.002286 mol

Therefore, Number of moles of reactans remained after the reaction,

nCH4 = 0.00714 - (0.002286/5) = 0.006683 mol

nO2 = 0.0397 - 0.002286 = 0.03741 mol

nNO = 0.00254 - 0.002286 = 0.000254 mol

Now, The number of moles of products formed,

nCO2 = nH2O = 0.0004572 mol

nNO2 = 0.002286 mol

nOH = 0.0009144 mol

Part A:

The partial pressures of reactants,

PCH4 = (0.006683*0.0821*275) / 2.0 = 0.075 atm

PO2 = (0.03741*0.0821*275) / 2.0 = 0.42 atm

PNO = (0.000254*0.0821*275) / 2.0 = 0.0029 atm

Total pressure of reactants = 0.075 + 0.42 + 0.0029 = 0.4979 atm

Part B:

PCO2 = (0.0004572*0.0821*275) / 2.0 = 0.0052 atm

PH2O = (0.0004572*0.0821*275) / 2.0 = 0.0052 atm

PNO2 = (0.002286*0.0821*275) / 2.0 = 0.026 atm

POH = (0.0009144*0.0821*275) / 2.0 = 0.010 atm

Total pressure of products = 0.0052 + 0.0052 + 0.026 + 0.010 = 0.0464 atm

Part C:

Total prssure in the flask = 0.4979 + 0.0464 = 0.5443 atm

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