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Thermochemistry problems

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nemzy:
1) 1 Cu2O(s)+ (1/2)O2(g) -> 2CuO(s) Change in H = -144 kj
2 Cu2O(s) -> Cu(s) + CuO(s) Change in H = +11
Calculate the standard enthalpy of formation of CuO

This is what i did, i reversed the 2nd equation so i get Cu(s) + CuO(s) -> Cu2O(s) by applying hess law, and Change in H becomes -11, then i added both equation 1 and 2 and got Cu(s)+(1/2)O2 -> CuO(s) after balancing, and the standard enthalpy becomes -144 kj - 11 kj which equals -155 kj.. Did i do this problem correctly?

The following questions i got wrong and i have no idea why...

1) Calculate the S*net for the following reaction, Given the S* vales: Na(s)=51.4 e.u, H20 (l) = 69.9 e.u , Na+ (aq) = 60.2 e.u., OH- (aq) = -10.5 e.u. H2(g) = 130.6
Reaction = Na(s) + H20(l) <---> Na+(aq) +OH- + H2(g)

This is how i solved it.. After balancing the equation, i got Na(s)+2H2O(l)<----> Na+(aq) +2OH- +H2(g)...And S*net = products - reactans, and after plugging it all in i got (60.2-2(10.5)+130.6) - (51.4 +2(69.9) which is -21.4 e.u. However, my answer was wrong and the answer was -12.6 e.u. How the heck did the professor get -12.6 ??

2) For the process:
benzene(l) --(1 atm)--> benzene (g), Change in Enthalpy = 30.5 kj/mol
Change in S*vap=86.4 J/mol*K

Assuming these vales are independent of T, what is the normal boiling point of benzene?

This is how i solved it, since it is in 1 atm , it is simply Change in enthalpy - T*change in entropy= Change in G (free energy) = 0

0 = Change in H - T*Change in S

After plugging it in.. 0 = 30,500 J/mol*k - T(86.4 J/mol*k) = 0

Solving for T you get 353degree Celcius..BUt that is not the answer, the answer is 80 degree celcius.. Where did i go wrong?

Donaldson Tan:

--- Quote from: nemzy on April 18, 2004, 09:02:35 PM ---Solving for T you get 353degree Celcius..BUt that is not the answer, the answer is 80 degree celcius.. Where did i go wrong?

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Solving T u get 353 Kelvin, not oC. 353K is equivalent to 80oC

Donaldson Tan:

--- Quote from: nemzy on April 18, 2004, 09:02:35 PM ---1) 1 Cu2O(s)+ (1/2)O2(g) -> 2CuO(s) Change in H = -144 kj
2 Cu2O(s) -> Cu(s) + CuO(s) Change in H = +11
Calculate the standard enthalpy of formation of CuO

This is what i did, i reversed the 2nd equation so i get Cu(s) + CuO(s) -> Cu2O(s) by applying hess law, and Change in H becomes -11, then i added both equation 1 and 2 and got Cu(s)+(1/2)O2 -> CuO(s) after balancing, and the standard enthalpy becomes -144 kj - 11 kj which equals -155 kj.. Did i do this problem correctly?

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