Question

1.) Calculate delta rH for the freezing of water at one bar and -10 degrees C. The heat capacities are H2O(s)=36.9 JK-1mol-1 and H2O(l)=74.5JK-1mol-1. delta H for freezing of water at 0 degree C is -6602Jmol-1 (assume C is independent of temperature).

2.) Assume the hot brick gives off 500J total. Calculate q,w, and delta U and delta H for gases in panel A and panel B respectively? (assume that gas in panel A experiences a change in volume of 1.5L during the process).

3.)At a very low temperature, most of solid its heat capacity has a dependence on temperature can be written as Cp=aT^3, a is a constant. Derive an expression for the change of enthalpy delta H from T=0 to a temperature T where the above relationship for Cp is valid.

4.) From the following data at 298.15K, calculate the standard enthalpy of formation of FeO(s) and Fe2O3(s).

Fe2O3(s)+3C(graphite)---->2Fe(s)+3CO(g… delta rH (kJmol-1)= 492.6
FeO(s)+C(graphite)---->Fe(s)+CO(g) delta rH=155.8
C(graphite)+O2(g))---->CO2(g) delta rH= -393.51
CO(g)+1/2O2(g)---->CO2(g) delta rH= -282.98

5.) An ice cube weighing 18g is removed from a freezer where it has been -20 degree C. (a) How much heat is required to warm the ice cube to ) degree C without it melting? (b) How much additional heat is required to melt it? (c) Suppose the ice cube was placed initially in a 180g sample of liquid water at +20 degree C in an insulated (thermally isolated) container. Describe the final state when the system has reached equilibrium. (delta H for water is 6.01KJ/mol. Heat capacity for water liquid is Cpm = 75.291 JK-mol-1 and is 70.391 JK-1mol-1 for ice).

6.)The constant pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression Cp/(JK-1)= 20.17+0.3665 (T/K). Calculate q, w, delta U, and delta H when the temp. is raised from 25 degree C to 200 degree C (a) at constant pressure (b) at constant volume.

7.) The constant pressure heat capacity of a sample of a perfect gas was found to vary with temperature according to the expression Cp/(JK-1)= 20.17+0.4001 (T/K). Calculate q, w, delta U, and delta H when the temp. is raised from 0 degree C to 100 degree C (a) at constant pressure (b) at constant volum

Answer 1

I'm going to start with the easiest ones and see how far I get.

#5. 18 g is one mole. (a) I guess you meant to 0 degrees C? So, 70.391 x 20 = 1408 J (b) 6.01 kJ (c) 180 g is 10 moles. It would take about 15kJ to cool all of the warmer water down to the freezing point, and that exceeds the sum of the answers from (a) and (b); therefore, the final state will be all liquid, and its temperature may be calculated as follows: 1408 + 6010 + 75.291 T = (10) (75.291) (20 - T) = 15058.2 - 752.91 T 828.2 T = 7640.2 => T = 9.2 C Final state is all liquid at 9.2 C

#3. delta-H = integral from 0 to T1 of Cp dT = integral from 0 to T1 of aT^3 dT = (aT^4)/4 The constant of integration will be zero because we are still considering a "delta-H" and don't need to worry about whether there was any enthalpy in the T=0 state.

#6 and #7. To find the Cv, which will be needed for delta-U and also for part (b) of each of these problems, one could use the relation Cv = Cp - R, but of course, not knowing the mass or moles of the sample, one does not know R either, and one does not know the number of atoms per molecule to use in such expressions as Cp = 1.4 Cv (for a diatomic molecule). It really seems as though there's a piece of missing information, especially because the ACTUAL Cp for monatomic gases such as argon, helium, and neon is right around 20 J/K PER MOLE. Making me think that the problem was meant to say this is a one-mole sample. (But this sample could just as well be 5/7 moles of diatomic gas, etc...)

Assuming monatomic gas, I obtain the following results: #6a. Delta-H = integral of Cp dT = 20.17 delta-T + 0.3665/2 times delta(T^2) = (20.17) (175) + 0.18325 (40000-625) = whatever [answer in Joules] Delta-U = integral of Cv dT. Since Cv = 3/5 Cp should be true at any temperature for a monatomic gas, the delta-U should be exactly 3/5 of the delta-H. Since dH = T dS + V dp, the Q in a constant-pressure change is just Delta-H. Q = delta-H and W = Q - delta-U = (2/5) delta-H #6b. At constant volume, W = 0, so Q = delta-U. As in #6a, Delta-H = integral of Cp dT and Delta-U = integral of Cv dT; the changes in enthalpy and internal energy come out just the same as in part (a), but the Q and W are different because of the different path taken. #7. Unless the different temperature dependency holds some clue about a different number of atoms per molecule, this problem appears to be the same as #6 with just some slightly different numerical inputs.

#1. I'm skipping this because I'm afraid "delta-rH" may mean something I'm not familiar with (it could be a typo on your part, but I'm not sure) #2. I'm skipping this one because "panel A and B" are not shown and I've no idea what they would show. #4. I'm skipping this one because I'm not very good at that type of thing.