Enthalpy Calculation Answers
1.	m = 100.0 g				q = n∆hvap = (m/M) ∆hvap
∆hvap = 40.1 kJ/mol H2O                                 	   = (100 g ÷ 18.02 g/mol)(40.1 kJ/mol H2O)
M = 18.02 g H2O/mol H2O		   	   = 22.5 kJ

2.	m = 500.0 g				q = n∆hvap = (500 g ÷ 62.02 g/mol)(58.8 kJ/mol)
∆hvap = 58.8 kJ/mol			   = 473.6 kJ
M = 62.02

3.	m = 1000 kg = 1 x 106 g			q = n∆hvap = (m/M) ∆hvap
∆hf = 6.03 kJ/mol			      	   = (1 x 106 g) ÷ (18.02 g/mol) (6.03 kJ/mol)
M = 18.02 g/mol			    	   = 3.35 x 105 kJ

4.	
System – NH2CONH2
    m = 10.0 g
    M = 60.07 g/mol
    ∆Hsol =    

     endothermic,  ∆H = +q

    ∆Hsol = ∆H/n = ∆HM/m
             = (2.322 kJ)(60.07 g/mol)/(10.0 g)
             = 13.95 kJ/mol	Surrounding – H2O
    m = 150.0 g = 0.150 kg
    c = 4.184 kJ/kgoC
    T1 = 20.4oC
    T2 = 16.7oC
    ∆T = 3.7oC

   q = mc∆T 
      = (0.150 kg)(4.184 kJ/kgoC)(3.7oC)
      = 2.322 kJ	

5.	  
6.	
(a) System – C2H6
      n = 5.00 mol C2H6
      ∆Hcomb = -1.56 MJ/mol

∆H = n∆Hcomb = (5 mol C2H6)(-1.56 MJ/mol)
       = -7.8 MJ	(b) System – C2H6
    m = 40.0 g
    M = 30.08 g/mol
      ∆Hcomb = -1.56 MJ/mol

∆H = n∆Hcomb 
       = (40.0 g C2H6) ÷ (30.08 g/mol)(-1.56 MJ/mol)
       = -2.30 MJ

7.	
System – AlCl3
    m = 40.0 g
    M = 133.33 g/mol
   ∆Hsol  = +14.8 kJ/mol

     endothermic, therefore ∆H = +q

   ∆H = n∆Hsol = + q = mc∆T
     (40.0 g)/(133.33 g/mol) = (0.150 kg)(4.184 kJ/kgoC)( ∆T)
      ∆T = 5.31oC	Surrounding – H2O
    m = 200.0 g = 0.200 kg
    c = 4.184 kJ/kgoC
    T1 = 20.4oC
    T2 = 
    ∆T =

∆T = T2 – T1 = T2 – 20.4oC  = 5.31oC
T2 = 19.69oC	
8.	  
System – C10H22
   m = 
   M = 144.52 g/mol
   ∆Hcomb = - 6.78 MJ/mol = 6.78 x 103 kJ/mol

    exothermic, therefore ∆H = -q	Surrounding – H2O
    m = 500.0 g = 0.500 kg
    c = 4.184 kJ/kgoC
    T1 = 20oC
    T2 = 55.5oC
    ∆T =35.0oC
	
                                                    n∆Hcomb = ∆H = -q
            m ÷ (144.52 g/mol)(-6.78 x 103) = (0.500 kg)(4.184 kJ/kgoC)(35oC)
       
     m = 1.56 g	

9.	
isolated system; 
no heat lost or gained to the calorimeter; 
for dilute solutions assume they have the specific heat capacity and density of pure water

10.	(a) ∆Hvap		(b) ∆Hsublimation 	(c) ∆Hsolubility 	(d) ∆Hcombustion 	(e) ∆Hneutralization 

11.	accuracy of the balance and the thermometer; determine how close you are to the true values and any errors made here will carry through all calculations

12.	
(a)	∆Hdilute = -10.88 kJ/mol
(b)	∆T would be to small, and therefore q would be less than expected resulting in a ∆Hdilut that is less than the expected value

13.	∆Hrxn = -25.98 kJ/mol

14.	∆H = -242.06 kJ

15.	∆Hneut = -64.11 kJ/mol