Entrop and enthalpy Watch

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Report Thread starter 2 years ago
I need some help... Completed most of the questions but stuck on these??*

The enthalpy of formation of water as steam is likely to be which of the following if the enthalpy of formation of water under standard conditions is -286 kJ/mole

*2. The entropy change for the reaction Al + Fe2O3 --- Al2O3 + Fe is?if the S of Al = 28 J/mol.k; Fe2O3 = 90 J/mol.k; Al2O3 = 51 J/mol.k; Fe = 27 J/mol.k

*3. The entropy change for the reaction 2Al + 3H2O --- Al2O3 + 3H2 is if the S values of Al = 28; H2 O(g) = 190; Al2O3 = 51; H2 = 130

*4. The entropy change for the reaction H2S + Cu -- CuS + H2 is?if the values of H2S = 206; Cu = 33; CuS = 66; H2 =130

*5. The entropy change of the surroundings for reaction that has an enthalpy change of reaction of -100 kJ/mole at 300K is?

Any help would be greatly appreciated....
Badges: 11
Report 2 years ago
The first question requires more data to work out the answer. Vapourisation of a liquid to a gas is endothermic (breaking Hydrogen bonds in this case) so you would expect formation of the gas to be less exothermic than formation of the liquid, so that rules out -310KJ/mol. After that I can't see how you would work out which of the other three values it is without more numbers. Googling the enthalpy of vapourisation for water (H2O(l)--> H2O(g)) gives +40.66KJ/mol (this value is given for the standard boiling point ie 373.15K (100degrees C). Now consider the process of forming the liquid and then evaporating, the enthalpy change would be -286+40.66= -245.34 KJ/mol. So the answer from the list is -242KJ/mol.(The difference between these answers arises because formation of the vapour is the value for forming the vapour at 298K, whereas the vapour in the calculation above is being formed at 373.15K and enthalpy value vary slightly with temperature. There are ways of accounting for this using more advanced thermodynamics, but that's well above anything expected to be understood at A level)The next three questions are thankfully more straightforward!The only thing required is the fact thatentropy change= total entropy of products - total entropy of reactantsremembering to multiply each species by the number of moles appearing in the reaction equation So question 2entropy of products = 1mole*51+1mole*27 = 78 J/(K.mol) entropy of reactants = 1mole*90+1mole*28 = 118 J/(K.mol) entropy change= 78-118 = -40J/(K.mol)questions 3 and 4 should hopefully be doable in exactly the same way The last question is interesting, and there are a few ways to think about it. The equation you need is as followsdeltaS(surroundings) = -deltaH(system)/ T(surroundings)so deltaS(surroundings) = -(-100KJ)/mol /300K = 0.3333 KJ(K.mol) = 333.33 J(K.mol) (We can reach this equation via several routes but this source http://chemwiki.ucdavis.edu/Core/Phy...e_Surroundings gets their directly from the second law of thermodynamics, this is above A level. (note, this source works in terms of heat (q) but this is equivalent to enthalpy when pressure is constant (something you can assume)).Anyway, this is my first post, hopefully it makes sense, I've tried to give some extra information, hopefully I haven't over complicated!

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