Question

Decomposition of methane in air can be increased in the presence of nitric oxide to form carbon dioxide, water, nitrogen dioxide, and OH where all species are gases. In order to measure the kinetics, a chemist placed 100 mL of methane at STP conditions into a flask with 700 mL of oxygen at STP and 40 mL of nitric oxide at STP.

A) Find the limiting reactant

B) The 3.0 L flask is held at 1.60 C until the reaction is 80% complete. Calculate the partial pressures of all reactants and products in the flask.

Answer 1

Part a

The balanced reaction

CH4 + 3O2 + NO = CO2 + H2O + NO2 + 2OH

Volume at STP Vs = 22.4 L/mol

Volume of CH4 V1 = 100 mL x 1L/1000 mL = 0.100 L

Moles of CH4 = V1/Vs = (0.100L)/(22.4L/mol) = 4.46*10^-3 mol

Moles of O2 = volume of O2/Vs

= (0.700L)/(22.4L/mol) = 0.03125 mol

Moles of NO = volume of NO/Vs

= (0.040L)/(22.4L/mol) = 1.785*10^-3 mol

From the stoichiometry of the reaction

1 mol CH4 reacts with = 1 mol NO = 3 mol O2

4.46*10^-3 mol CH4 reacts with = 4.46*10^-3 mol NO = (3*4.46*10^-3 =) 0.01338 mol O2

We have more moles(0.03125) of O2 than required (0.01338)

But we have less moles (1.785*10^-3) of NO than required (4.46*10^-3)

Limiting reactant = NO

Part b

Volume of flask V = 3.0 L

Temperature T = 1.60°C + 273.15K = 274.75 K

Conversion = 0.80

Moles of NO react = 1.785*10^-3 * 0.80 = 0.001428 mol

Moles of NO unreacted = 1.785*10^-3 * (1-0.80) = 0.000357 mol

1 mol NO reacts with = 3 mol O2

Moles of O2 reacted = 3*moles of NO reacted

= 3*0.001428 = 0.004284 mol

Moles of O2 unreacted = Initial Moles of O2 - moles of O2 reacted

= 0.03125 - 0.004284

= 0.026966 mol

Moles of CH4 reacted = moles of NO reacted = 0.001428 mol

Moles of CH4 unreacted = 0.0046 - 0.001428 = 0.003032 mol

Moles of CO2 formed = moles of NO reacted = 0.001428 mol

Moles of H2O formed = moles of NO reacted = 0.001428 mol

Moles of NO2 formed = moles of NO reacted = 0.001428 mol

Moles of OH formed = 2*moles of NO reacted

= 2*0.001428 mol

= 0.002856 mol

the flask contains all unreacted reactants and all formed products

Moles of CH4 = 0.003032 mol

Partial pressure of CH4 = nRT/V

= 0.003032 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.003032 mol x 7.519 atm/mol

= 0.0228 atm

Moles of O2 = 0.026966 mol

Partial pressure of O2 = nRT/V

= 0.026966 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.026966 mol x 7.519 atm/mol

= 0.2028 atm

Moles of NO = 0.000357 mol

Partial pressure of NO = nRT/V

= 0.000357 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.000357 mol x 7.519 atm/mol

= 0.00268 atm

Moles of CO2 = 0.001428 mol

Partial pressure of CO2 = nRT/V

= 0.001428 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.001428 mol x 7.519 atm/mol

= 0.0107 atm

Moles of H2O = 0.001428 mol

Partial pressure of H2O = nRT/V

= 0.001428 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.001428 mol x 7.519 atm/mol

= 0.0107 atm

Moles of NO2 = 0.001428 mol

Partial pressure of NO2 = nRT/V

= 0.001428 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.001428 mol x 7.519 atm/mol

= 0.0107 atm

Moles of OH = 0.002856 mol

Partial pressure of OH = nRT/V

= 0.002856 mol x 0.0821 L-atm/mol-K x 274.75 K / 3.0 L

= 0.002856 mol x 7.519 atm/mol

= 0.0215 atm