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 145 mL of methane at STP, 875 mL of oxygen at STP, and 56.0 mL of NO at STP in a 1.8 L flask. The reaction is allowed to stand for several weeks at 275 K

Part A)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?

Part B)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?

Part C) What is the total pressure in the flask?

Explanation / Answer

Solution :-

CH4(g)+5O2(g)+5NO(g) ------- > CO2(g)+H2O(g)+5NO2(g)+2OH(g)

Lets calculate the moles of the each reactant in the given volume at STP

Moles of methane = 0.145 L * 1 mol / 22.4 L = 0.006473 mol

Moles of O2 = 0.875 L * 1 mol / 22.4 L = 0.03906 mol

Moles of NO = 0.056 L *1 mol / 22.4 L = 0.0025 mol NO

Now lets calculate the moles of the CH4 and O2 needed to react with 0.0025 mol NO

0.0025 mol NO * 1 mol CH4 / 5 mol NO = 0.0005 mol CH4

0.0025 mol NO * 5 mol O2 / 5 mol NO = 0.0025 mol O2

So the NO is the limiting reactant

Therefore lets find the moles of the each reactant reacted when the percent yield is 92 %

CH4 = 0.0005 mol * 92 % / 100 % = 0.00046 mol

O2 = 0.0025 mol * 92 % / 100 % = 0.0023 mol O2

NO = 0.0025 mol *92%/100% = 0.0023 mol

Now lets calculate the moles of the each reactant remain after the reaction

Moles of CH4 remain = 0.006473 mol – 0.00046 mol = 0.006013 mol

Moles of O2 remain = 0.039 mol – 0.0023 mol = 0.0367 mol

Moles of NO remain = 0.0025 – 0.0023 = 0.0002 mol

Now lets calculate the moles of the products produced

Using the mole ratio of the NO

0.0023 mol NO * 1 mol CO2/5 mol NO = 0.00046 mol CO2

0.0023 mol NO * 1 mol H2O / 5 mol NO = 0.00046 mol H2O

0.0023 mol NO * 5 mol NO2 / 5 mol NO = 0.0023 mol NO2

0.0023 mol NO * 2 mol OH / 5 mol NO = 0.00092 mol OH

Now lets calculate the total moles in the flask

Total moles = sum of reactant moles + sum of product moles

Total moles = 0.006013 + 0.0367 +0.0002 +0.00046+0.00046+0.0023+0.00092

                      = 0.047053 mol

Now lets calculate the total pressure in the flaks after the reaction

PV= nRT

P= nRT/V

P= 0.047053 mol * 0.08206 L atm per mol K * 275 K / 1.8 L

P= 0.5899 atm

Now lets calculate the partial pressure of each reactant

[CH4] =[0.006013/0.047053]*0.5899 atm = 0.0754 atm

[O2] =[0.0367/0.047053]*0.5899 atm = 0.4601 atm

[NO] = [0.0002/0.047053] * 0.5899 atm = 0.002507 atm

Now lets calculate the partial pressure of each product

[CO2]= [0.00046/0.047053]*0.5899 atm = 0.005767 atm

[H2O] = [0.00046/0.047053]*0.5899 atm = 0.005767 atm

[NO2] = [0.0023 / 0.047053] *0.5899 atm = 0.0288 atm

[OH] = [0.00092/0.047053]*0.5899 atm = 0.0115 atm

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