Problem 9.128 An empty 4.25 L steel vessel is filled with 1.15 atm of CH4( g ) a

Question

Problem 9.128

An empty 4.25 L steel vessel is filled with 1.15 atm of CH4(g) and 4.60 atm of O2(g) at 300 C. A spark causes the CH4 to burn completely, according to the equation:
CH4(g)+2O2(g)CO2(g)+2H2O(g) H = -802kJ

Part A

What mass of CO2(g) is produced in the reaction?

mCO2=4.57

Part B

What is the final temperature inside the vessel after combustion, assuming that the steel vessel has a mass of 12.475 kg , the mixture of gases has an average molar heat capacity of 21J/(molC), and the heat capacity of steel is 0.499J/(gC)?

Part C

What is the partial pressure of CO2(g) in the vessel after combustion?

mCO2=4.57

Part B

What is the final temperature inside the vessel after combustion, assuming that the steel vessel has a mass of 12.475 kg , the mixture of gases has an average molar heat capacity of 21J/(molC), and the heat capacity of steel is 0.499J/(gC)?

Part C

What is the partial pressure of CO2(g) in the vessel after combustion?

Explanation / Answer

1.

Moles of reactants can be calculated from the gas law equation, n= PV/RT, Pm in atm, V in L , R =0.0821 l.atm/mole.K , T in K

For CH4, moles= 1.15*4.25/ (0.0821* (300+273.15)=0.102 moles.

For O2, moles = 4.6*4.25/(0.0821*(300+273.15)=0.42

The reaction between methane and Oxygen is

CH4+2O2---->CO2 +2H2O

1 mole of CH4 requires 2 mole of Oxygen . So molar ratio of reactants is =1 :2

Actual ratio given = 0.102: 0.42 = 1:4.11. So oxygen is excess reactant and limiting reactant is CH4. So moles of CO2 formed= 0.102moles.

Mass of CO2= 0.102*44=4.5 gms

2.

The heat generated is taken by the products, the vessel

Let the final temperature be T

Gases leaving : CO2= 0.102 moles =4.5 gms, moles of H2O = 2*0.102= 0.204moles. mass of water= 0.204*18= 3.672 gms oxygen remaining ( since 1 mole of CH4 requires 2 mole of oxygen) = 0.42-2*0.102= 0.216 moles. Mass of oxygen= 0.216*32= 6.9 gms

Moles of products= 0.204+0.102 +0.216=0.522 moles

So heat taken by products= 0.522*21*(T-573.15)=11*(T-573.15)

Heat taken by vessel = 12.475*0.499*(T-573.15) =6.23*(T-573.15)

Total heat taken = 11*(T-573.15)+6.23*(T-573.15)= 17.23*(T-573.15)

This is equal to 802*1000 Joules/moles of CH4. Heat liberated= 802*1000*0.102

17.23*(T-573.15)= 802*1000*0.102

T-573.15= 802*1000*0.102/17.23, T= 5021K

After mixing, the pressure needs to be calculated from

Total number of moles = 0.522 moles, V= 4.25 L.

P= nRT/V= 0.522*5021*0.0821/4.25=50.63 atm

Partial pressure of CO2= mole fraction* total pressure = (0.102/0.522)*50.63= 9.89 atm

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