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posted by  geniva on 11/20/2008 11:42:57 AM  |  status: Live  

2nd Part

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General Chemistry N/A N/A N/A
Question Details:
If the Hindenburg was 2.2 x 10 8 L of H2(g) at 1 atm and 25 degrees celsius, Calculate the ΔH if 100% of the H2(g) burned. 
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posted by Werner on 11/21/2008 4:45:54 AM  |  status: Live
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geniva's comment:
"You are so smart and explain the answers in detail so I can understand them."
Response Details:
H2(g) + 1/2 O2(g)  --------> H2O(l)  ΔH = -286kJ/mole  
    The above equation shows 286 kJ heat is liberated when 1 mole of H2(g)  is burned.
  Given data
  The volume of H2(g), V = 2.2*108 L 
   Pressure of H2(g) , P = 1 atm 
   Temperature, T = 25 deg C
                            = 298 K
   According to the ideal gas equation
                       PV = nRT
 ∴ Number of moles of H2(g) , n = PV/RT
                                                   = (1 atm)( 2.2*108 L ) /(0.0821 L atm/mol.K)(298K)
                                                   = 8.992*106 mol
    ΔHrxn when 2.2*108 L of H2(g) is burned = {-286kJ/mole H2(g)}* 8.992*106 mol H2(g)    
                                                                     = -2.571*109 kJ
                                       
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