please answer the question number 3 For the reaction: COCI COdgo C Kp 8 x 10\" a

Question

please answer the question number 3

For the reaction: COCI COdgo C Kp 8 x 10" atm at 100 and AS (3730K) 125 JPK. a) Determine the degree of dissociation of COC12 at 1000C and 2 atm pressure b) Calculate AH (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 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: Sos) 2 Co SO 2C. (g) 2(g) (s) 5. Helium has two liquid phases (I and II) which form a triple point with the gas at 0.075 bar (1 bar i atm) and 2.17 0K. The boiling point of liquid He is 4.22 K. Using the attached graph, find an approximate value for the heat of vaporization of liquid He I in units of R. 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 for Ice I is -100a ile that for Ice III is +300 atm/C. The density of Ice I 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 AHvT 8 j/deg-mole, where Tb is the boiling point and AHv 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 normal boiling point (in 0K) b. The normal boiling point of a liquid is 120 0C. Calculate the vapor pressure of the liquid at 121 0C. c. The vapor pressure of acetonitrile is changing at the rate of 0.030 atm/deg 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

Answer to Q3a)

COCl2(g) <------> CO(g) + Cl2(g)

If is the degree of dissociation at 373 K, then using ICE table, we can find the concentrations at the equilibrium.

[COCl2]

[CO]

[Cl2]

Initial

1

0

0

Change

-

+

+

Equilibrium

1-

Kp = (CO) (Cl2) / (COCl2)

Where,

(CO) = partial pressure of CO = CO Ptotal

CO= The mole fraction of CO = nCO : ntotal

total number of moles (ntotal ) = 1- + + = 1+

Total pressure (Ptotal) = 2 atm (given)

(CO) = • 2 / 1+

Similarly, (Cl2) = partial pressure of Cl2 = Cl2 Ptotal

= • 2 / 1+

(COCl2) = partial pressure of COCl2 = COCl2 Ptotal

= (1- ) • 2 / 1+

Kp = (CO) (Cl2) / (COCl2)

= (2 / 1+ ) X (2 / 1+ ) / (2(1- ) / 1+ )

= (2 / 1+ ) X (2 / 1+ ) X (2(1- ) / 1+ )

= 22 / (1- ) (1+ )

= 22 / (1- )2

8 X 10-9 = 22 / (1- )2

Since Kp is very small, will be very small. Therefore 1- = 1

8 X 10-9 = 22

2 = 8 X 10-9 / 2

2 = 4 X 10-9

= 6.3 X 10-5 M

Answer to Q3b)

The Gibbs free energy change for this reaction is given as:

G0 = [G0f (CO(g)) + G0f (Cl2(g))] - G0f (COCl2(g))

The G0f values are given in the thermodynamics table

G0 = [(-137.2) + (0)] – (-204.6)

= 67.4 kJ

We know,

G0 = H0 - TS0

H0 = G0 + TS0

= 67.4 kJ + 373 K X 0.125 kJ K-1

= 67.4 kJ + 46.6 kJ

H0 = 114 kJ

Answer to Q3c)

Previously we calculated,

Kp = 2 Ptotal / (1- )2

When = 0.001, P = 2 atm

Kp = (0.001)2 2/ (1- 0.001)2

= 0.000002 / 0.998

= 2 X 10-6

We know,

G = G0 +RT lnQ

At equilibrium G = 0 and Q = Kp

G0 = -RT ln Kp

T= -G0 / R ln Kp

= -67.4 X 103 J mol-1/ 8.314 J mol-1 K-1 • ln 2 X 10-6

= -67.4 X 103 / 8.314 K-1 • (-13.12)

T = 617.9 K

[COCl2]

[CO]

[Cl2]

Initial

1

0

0

Change

-

+

+

Equilibrium

1-

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