A gas mixture #3.67L# in volume contain #C_2H_4# and #CH_4# in proportion of #2:1# by moles and is at #25^oC# and 1atm. If the #DeltaH_c(C_2H_4)# and #DeltaH_c(CH_4)# are #-"1400 kJ/mol"# and #-"900 kJ/mol"# find heat evolved on burning this mixture?

A)#20.91 kJ#
B)#50.88 kJ#
C)#185 kJ#
D)#160 kJ#

1 Answer
Truong-Son N.
Jan 6, 2018

I got #"185 kJ"#, though I can only tell you #"200 kJ"#.

Well, heat is additive, so you just need to find the mols of each gas in the mixture, and you can then add up the enthalpies of combusting each gas individually.

We assume each gas is ideal. Then the ideal gas law applies.

#PV = nRT#

  • #P# is the pressure in #"atm"#.
  • #V# is volume in #"L"#.
  • #n# is the mols of an ideal gas.
  • #R = "0.082057 L"cdot"atm/mol"cdot"K"# is the universal gas constant.
  • #T# is temperature in #"K"#.

The total mols of gas is:

#n = (PV)/(RT) = ("1 atm" cdot "3.67 L")/("0.082057 L"cdot"atm/mol"cdot"K" cdot "298.15 K")#

#=# #"0.15 mols gas"#

Since the gas mixture is #"2 mols C"_2"H"_4:"1 mol CH"_4#, there are #"0.10 mols C"_2"H"_4# and #"0.05 mols CH"_4#.

As a result, the heat produced from combusting each one is:

#|DeltaH_C("C"_2"H"_4)| = |-"1400 kJ"/"mol"| xx "0.10 mols C"_2"H"_4#

#=# #+"140 kJ"#

#|DeltaH_C("CH"_4)| = |-"900 kJ"/"mol"| xx "0.05 mols CH"_4#

#=# #+"45.0 kJ"#

This gives in total, #color(blue)(DeltaH_C = "185 kJ")#. But you only get 1 significant figure based on your pressure.

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