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I need help with SL ThermoChem


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1) 870 J of heat is required to vaporize 4.5 g liquid bromine (Br2). Whta is the molar heat of vaporization of Br2 in kJ/mol? The answer is 31. 

 

2) The evaporation of 1.00g of ethanol is accompanied by an enthalpy change of 879 J. What the enthalpy for the following change in state?

 

CH3CH2OH (g) -> CH3CH2OH (l)

Edited by frank!e
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1) 870 J of heat is required to vaporize 4.5 g liquid bromine (Br2). Whta is the molar heat of vaporization of Br2 in kJ/mol? The answer is 31. 

 

2) The evaporation of 1.00g of ethanol is accompanied by an enthalpy change of 879 J. What the enthalpy for the following change in state?

 

CH3CH2OH (g) -> CH3CH2OH (l)

 

Look, the tactic to tackle this type of questions is to use cross calculation. For question 1, the first step is to realize that the question asks for kJ/mol, which means that you should first calculate how many moles 4.5g of Br2 contains. That is a simple calculation and should give you 0.028125mol. Now the next step is to rewrite the question is a good format, like this:

 

0.028125 mol ==>  0.870 kJ  (in words, 0.028125 mol of Br2 requires 0.870 kJ)

         1     mol  ==>     ?    kJ  (so how many kilojoules does 1 mole of Br2 need?)

 

Now, if you perform a cross calculation (i.e. cross multiply the 2 lines above in order to get the value for the question mark), you should get: 0.870 * 1 / 0.028125 = 31 kJ/mol. Applying the same tactic for question 2 should give you:

 

0.021739 mol ==>  0.879 kJ  (in words, 0.021739 mol of ethanol requires 0.879 kJ)

       1      mol  ==>     ?    kJ  (so how many kilojoules does 1 mole of ethanol need?)

 

Cross calculation should give you 40.4 kJ/mol. However, there's a little twist in the second question. If you look at the chemical equation, you can see that ethanol changes from the gas to the liquid phase, which means that it's an exothermic reaction. Thus the final answer must be -40.4 kJ/mol (i.e. negative enthalpy change)

Edited by Vioh
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