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 165 mL of methane at STP, 865 mL of oxygen at STP, and 57.5 mL of NO at STP in a 2.2 −L flask. The reaction is allowed to stand for several weeks at 275 K.

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? (answer as follows: PCH4,PO2,PNO)

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? (answer as follows PCO2,PH2O,PNO2,POH)

C) What is the total pressure in the flask?

Answer 1

1 mole of a gas at STP has 22400 ml volume.

So, No. of moles of CH₄ taken = 165/22400 = 1.12 x 10^-6

No. of moles of O₂ taken = 865/22400 = 3.86 x 10^-2

No. of moles of NO taken = 57.5/22400 = 2.56 x 10^-3

We have following balanced equation

CH_4(g)+5O_2(g)+5NO_(g)→CO_2(g)+H₂O_(g)+5NO_2(g)+2OH_(g)

Now, 1 mole of CH₄ will require 5 moles of O₂ and 5 moles of NO.

So, 1.12 x 10^-6 mole of CH₄ will require 5*1.12 x 10^-6 moles of O₂ and 5*1.12 x 10^-6 moles of NO.

So, 1.12 x 10^-6 mole of CH₄ will require 5.60 x 10^-6 moles of O₂ and 5.60x 10^-6 moles of NO.

So, CH₄ is Limiting Reagent as it will get consumed prior to all other reactants.

(A) Now, when 90% of CH₄ will get consumed then moles of CH₄ reacted = 90% of 1.12 x 10^-6

   = 1.008 x 10^-6

Now, when 90% of CH₄ will get consumed then moles of CH₄ left unreacted = 10% of 1.12 x 10^-6

   = 1.12 x 10^-7

So, when 1.008 x 10^-6 moles of CH₄ reacted then moles of O₂ and NO reacted will be 5 times this value i.e. 5.04 x 10^-6 moles.

At 90% completion of reaction, No. of moles of CH₄ left = 1.12 x 10^-7

So, Partial Pressure due to CH₄ , P_CH4 = 1.12 x 10^-7 x 0.0821 x 275 / 2.2

P_CH4 = 1.149 x 10^-7 atm

Similarly,

No. of moles of O₂ left =  3.86 x 10^-2 - 5.04 x 10^-6 = 0.03859496

So, Partial Pressure due to O₂ , P_O2 = 0.03859496 x 0.0821 x 275 / 2.2

P_O2 = 0.396 atm

No. of moles of NOleft =  2.56 x 10^-3 - 5.04 x 10^-6 = 0.00255496

So, Partial Pressure due to NO , P_NO​ =  0.00255496 x 0.0821 x 275 / 2.2

  P_NO = 0.026 atm

(B) Now, when 90% of CH₄ will get consumed then moles of CH₄ reacted = 90% of 1.12 x 10^-6

   = 1.008 x 10^-6

So by given balanced reaction, when 1.008 x 10^-6 moles of CH₄ reacted then moles of CO₂, H₂O, NO₂ and OH formed will be 1.008 x 10^-6, 1.008 x 10^-6, 5*1.008 x 10^-6 and 2*1.008 x 10^-6 moles respectively.

So, Partial Pressure due to CO₂ , P_CO2 = 1.008 x 10^-6 x 0.0821 x 275 / 2.2

P_CO2 = 1.034 x 10^-5 atm

&, Partial Pressure due to H₂O , P_H2O =  1.008 x 10^-6 x 0.0821 x 275 / 2.2

P_H2O​ = 1.034 x 10^-5 atm

&, Partial Pressure due to NO₂ , P_NO2 = 5*1.008 x 10^-6 x 0.0821 x 275 / 2.2

P_NO2 = 5.17 x 10^-5 atm

&, Partial Pressure due to OH , P_OH = 2*1.008 x 10^-6 x 0.0821 x 275 / 2.2

P_OH = 2.068 x 10^-5 atm

(C) So, Total Pressure will be equal to sum of all the partial pressures

According to Dalton's law of Partial pressure P_total = p₁ +p₂ +.....

So, P_total = P_CH4 + P_O2 + P_NO + P_CO2 + P_H2O + P_NO2 + P_OH

= 1.149 x 10^-7 + 0.396 + 0.026 + 1.034 x 10^-5 + 1.034 x 10^-5 + 5.17 x 10^-5 + 2.068 x 10^-5

P_total = 0.4221 atm