Midterm Questions that i missed

[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?

 

[Zefram]

it is important to always double check your calculations and make sure you are using the correct values and units. In the first question, it seems like you may have used the wrong values for the entropy of Na+ and OH-. The correct values are 51.4 e.u. and 10.5 e.u., respectively. Using these values, the net entropy change is indeed -12.6 e.u.

In the second question, you have correctly set up the equation for ΔG, but the units you have used for ΔH are incorrect. The value given in the problem is in kJ/mol, so you need to convert it to J/mol before plugging it into the equation. This will give you a value of 30,500 J/mol, which when solved for T will give the correct boiling point of 80 °C.

It is important to be careful with units and double check your calculations to avoid mistakes. Additionally, make sure you are using the correct values given in the problem and not mixing them up with other values.

 

[ravenprp]

For the first question, you did not take into account the coefficients of the balanced equation when calculating the S*net. The correct equation would be Na(s) + 2H2O(l) <---> Na+(aq) + 2OH-(aq) + H2(g). Then, the S*net would be (60.2 + 2(-10.5) + 130.6) - (51.4 + 2(69.9)) = -12.6 e.u.

For the second question, you should be using the units of kJ/mol*K for both enthalpy and entropy. So the correct equation would be 0 = 30.5 kJ/mol*K - T(0.0864 kJ/mol*K) = 0. Solving for T, you get 353 K, which is equivalent to 80 degrees Celsius.