Question
please answer the question number 6.
For the reaction: COCl2(g) CO(g) Cl2(g) Kp 8 x 109 atm at 1000C, and AS (3730K) 125 JPK. a) Determine the degree of dissociation of CoCl2 at 1000C and 2 atm pressure b) Calculate AH0 (3730K) for the reaction c) Assuming heat capacities do not change appreciably, at what temperature would the degree of dissociation be 0.001 (still at 2 atm) 4. When two moles of Co are introduced into a vessel containing solid sulfur, the final equilibrium pressure is 1.03 atm. Determine Kp for the reaction: S(s) 2 CO(g) 2(g 2C 5. Helium has two liquid phases (I and II) which form a triple point with the gas at 0.075 bar (lbar 1 atm) and 2.17 %K. The boiling point of liquid He is 4.22 oK Using the attached graph, find an approximate value for the heat of vaporization of liquid He I in units of R. 6. Consider the phase diagram (attached) for water in the region of the triple point between the phases Ice I, Ice III, and liquid water. The slope for the melting curve is 0.917 gm/ml, while that for liquid water is 1.000 gm/ml. Estimate the density of Ice III. You must show your reasoning. Trouton's rule states that AHv/Tb 88 j/deg-mole, where Tb is the boiling point and AHy is the heat of vaporization. (This rule holds for almost all substances) a. For a liquid, which obeys the Clausius-Clapyron equation, calculate the vapor pressure of the liquid at a temperature equal to one-third of its b. The normal boiling point of a liquid is 120 0c. Calculate the vapor pressure of the liquid at 121 oC atm/deg c. The vapor pressure of acetonitrile is changing at the rate of 0.030 near its boiling point of 80 C. Calculate its heat of vaporization. A certain substance exists in two phases at equilibrium at temperature TI and pressure P1. One phase a is a crystalline solid while the other B is an amorphous
Explanation / Answer
The density of ice III can be calculated from clausius clapeyron equation: dP/dT= (H2-H1)/ T*(V2-V1)
where dP/dT is slope of the curve in phase diagram. Since we are dealing with the density, we substitute V= M/D in the above equation where M= molecular weight and D= density.
The melting curve of ice III and ice I meet at temperature T= -220C . Hence we solve the equation at the same temperature: for ice III: -100= (H1-Hliq)/T*(V1-Vliq) or T= -(H1-Hliq)/100*(V1-Vliq).....eqn 1
for ice III: 300= (H3-Hliq)/T*(V3-Vliq) or T= (H3-Hliq)/300*(V3-Vliq).....................eqn 2
Now at the meeting point of melting curves of ice I and ice III, difference in enthalpies become equal, hence (H1-Hliq)= (H3-Hliq) and eqn 1 & eqn 2 can be equalised as:
300*(V3-Vliq)= - 100*(V1-Vliq).
Putting V= M/D = 18.00/ D, we get : 3* (18/D3- 18/1) = -(18/0.917 -18/1)
3*(18/D3- 18) = -(19.63-18) or 3*18(1/D3 - 1)= -1.63
1/D3= 1 - 1.63/54 = 1- 0.03 = 0.97 or D3 (density of ice III) = 1.031 gm/ml
