Question

A 5.20 g sample of trans-1,2-diethylcyclopropane (formula C₇H₁₄), is mixed with 35.67 atm of O₂ in a 7.38×10^-1 L combustion chamber at 204.18 °C. The combustion reaction to CO₂ and H₂O is initiated and the vessel is cooled back to 204.18 °C. Report all answers to two decimal places in standard notation (i.e. 1.23 atm).


1. What is P_H2O (in atm) in the combustion chamber assuming the reaction goes to completion? Assume all water is water vapour and that the volume of the system does not change.

2. What is P_O2 (in atm) in the combustion chamber after the reaction goes to completion?

Answer 1

The combustion reaction between 1,2-diethylcyclopropane (C₇H₁₄ ) and Oxygen is given by

C7H14+ 10.5 O2 ---->7CO2 + 7 H2O

Molecular weights

C7H14= 7*12+14*1= 98 O2= 32, CO2= 12+32= 44 and H2O= 1*2+16=18

Moles of C7H14= 5.2/98=0.053061

Moles of oxygen (n) can be calculated from ideal gas law equation PV= nRT

P= 35.67 atm V= 7.38*10^-1 L, R= 0.08206L.atm/mole.K T= 204.18+273.15K=477.33 K

Moles of Oxygen (n) before combustion= PV/RT= 35.67*7.38*10^-1/ (0.08206*477.33)=0.672 moles of Oxygen

As per the reaction, 1 moles of C7H14 requires 10.5 moles of Oxygen for complete combustion to produce 7 moles of CO2 and 7 moles of H2O vapor.

0.053061 moles requires 10.5*0.053061=0.557 moles of Oxygen

Oxygen remaining after combustion = Oxygen supplied – Oxygen consumed= 0.672-0.557=0.115 moles

CO2 formed= 7*0.053061=0.371429 and H2O formed= 0.371429

Total moles (n2) after complete combustion= 0.115(O2)+ 0.371429 (CO2)+ 0.371429 (H2O)=0.857857

Temperature (T2) after combustion = 477.33 K

V2= 7.38/10 L= 0738L , R =0.08206L.atm/mole

From P2V2= n2RT2 P2 ( final pressure )= n2RT2/V2= 0.857857*0.08206*477.33/(0.738)=45.53 atm

After combustion

Partial pressure of H2O= moles fraction of Water vapour* total pressure (P2)

Mole fraction = moles of water vapour/ total moles = 0.371429/0.857857=0.4330

Partial pressure of water= 0.4330*45.53=19.614 atm

Partial pressure of Oxygen= (0.115/0.857857)*45.53=6.10 atm