nemzy Posted April 19, 2004 Share Posted April 19, 2004 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? Link to comment Share on other sites More sharing options...
Gampin Posted April 26, 2004 Share Posted April 26, 2004 2) Solving for T, you actually get 353 K. Then, 353 - 273 = 80 C. Link to comment Share on other sites More sharing options...
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