An empty 4.00 L steel vessel is filled with 1.013 bar CH_4(g) and 4.053 bar O_2(

Question

An empty 4.00 L steel vessel is filled with 1.013 bar CH_4(g) and 4.053 bar O_2(g) at 300 degreeC. A spark causes the CH_4 to burn completely according to the balanced equation: CH_4(g) + 2 O_2(g) rightarrow CO_2(g) + H_2O(g) delta_rH^o = -802 kJ What mass of CO_2 is produced in the reaction? The steel vessel has a mass of 14.500 kg and a heat capacity of 0.449 J g^-1 degreeC^-1. The mixture of gases has an average molar heat capacity of 21 J mol^-1 degreeC^-1. What is the final temperature inside the vessel? What is the partial pressure of CO_2(g) (in bar) in the vessel after combustion?

Explanation / Answer

Volume of vessel =4 L

Moles of CH4= PV/RT , P is in atm, V in L , R=0.08206 L.atm/mole.K T= 300+273.15= 573.15K

1.013 bar= 1.013*0.9865=1 atm

n= 1*4/(573.15*0.08206) =0.085 moles of CH4

Moles of O2= 4.053*0.9869*4/(0.08206*573.15)= 0.34 moles

CH4 is limiting reactant and oxygen is the excess reactant ( since the molar ratio of CH4 to O2 is 1:2) where as supplied ratio is 0.085:0.34 =1:4)

Moles of CO2 formed from the reaction = 0.085 moles of H2O formed= 0.085 moles of OXygen remaining = 0.34-0.085*2=0.17 moles

Mass of CO2 produced= 0.085* 44 ( molecular weight of CO2)= 3.74 gms

Total moles of product = 0.085+0.085+0.17= 0.34

heat gained = 802 Kj = heat gained by vessel + heat gained by gases

= 802*1000= 0.449*14500*(T-300)+ 0.34*21*(T-300)

802*1000= (T-300)*(0.449*14500+0.34*21)=6517.4*(T-300)

T-300= 802*1000/6517.4= 123

T= 300+123= 423 deg.c

From PV= nRT, the pressure of CO2 can be calculated

P= nRT/V= 0.085*0.08206*(423+273.15)/4= 1.21 atm= 1.21/0.9869 bar= 1.23 bar

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