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: CH_4(.g) + 5O_2(g) + 5NO(g)rightarrow CO_2(g) + H_2O(g) + 5NO_2(g) + 2 OH(g) Suppose that an atmospheric chemist combines 150 inL of methane at STP 885 niL of oxygen at STP, and 58.5 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

Initial moles of,

CH4 = 0.150 L/22.4 L = 0.0067 mols

O2 = 0.885 L/22.4 L = 0.04 mols

NO = 0.0585 L/22.4 L = 0.0026 mols

Limiting reagent

If all of CH4 is consumed, we would need = 5 x 0.0067 = 0.0335 mols of O2 and 0.0335 mols of NO

If all of NO is consumed we would need = 0.0026/5 = 0.0052 mols of CH4 and 0.0026 mols of O2

Since the moles of NO are less than required, this is the limiting reagent

initial molar concentration of,

[CH4] = 0.0067 mol/1.8 L = 0.0037 M

[O2] = 0.04 mol/1.8 L = 0.022 M

[NO] = 0.0026 mol/1.8 L = 0.0014 M

92% of NO is consumed, remaining [NO] = 0.0014 - 0.0014 x 0.92 = 0.0001 M

remaining [O2] = 0.022 - 0.0013 = 0.021 M

[CH4] = 0.0067 - 0.0013 = 0.0054 M

Partial pressures of each reactant at 275 K,

P[CH4] = 0.0054 x 1.8 = 0.0037 mol

P[O2] = 0.021 x 1.8 = 0.038 mol

P[NO] = 0.0001 x 1.8 = 0.00018 mol

Total moles = 0.042 mol

Total pressure = 0.042 x 0.08205 x 275/1.8 = 0.526 atm

Part A :

Partial pressure for gases at 275 K,

p[CH4] = 0.0037 x 0.526 = 0.0019 atm

p[O2] = 0.038 x 0.526 = 0.02 atm

p[NO] = 0.00018 x 0.526 = 9.47 x 10^-5 atm

Part B :

Partial pressures of products at 275 K

moles of products,

[CO2] = [H2O] = 0.0013 x 1.8/5 = 0.00047 mol

[NO2] = 0.0013 x 1.8 = 0.0023 mol

[OH] = 2 x 0.0013 x 1.8/5 = 0.00093 mol

Total moles = 0.0037 mol

Total pressure = 0.0037 x 0.08205 x 275/1.8 = 0.046 atm

p[CO2] = 0.00047 x 0.046 = 2.16 x 10^-5 atm

p[H2O] = 0.00047 x 0.046 = 2.16 x 10^-5 atm

p[NO2] = 0.0023 x 0.046 = 1.06 x 10^-4 atm

p[OH] = 0.00093 x 0.046 = 4.28 x 10^-5 atm

Part C :

Total pressure in the flask = 1.92 x 10^-4 atm

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