question number 3 Given the following in the gas phase: CH_3OH, CO, CO_2, H_2 O,

Question

question number 3

Given the following in the gas phase: CH_3OH, CO, CO_2, H_2 O, H_2. Find the transfer of components and degrees of freedom. Write down the reactions that are constrains on the number of components. Given the delta G degree (298 K) for the following species: CH_3OH: -166.3, CO, -137.2, CO_2, -194.4, H_2O, -228.6, all in KJ. Find K for each reaction that is used as a constraint. Suppose the temperature drops and H_2O freezes, and CH_2OH condenses to a liquid. What are the new values for components, degreed of freedom, and phases? Starting from one mole of N_2 and one mole of O-2, find the equilibrium ratio of N_2/NO_3 t 298k. (Neglect any other possible nitrogen - oxygen reaction) (Given N_2 + 2O_2 rightarrow 2NO_2) Find the same ratio at 350 K; assume delta C_p is small enough to neglect it. Given: Delta_T G degree (NO_2, 298 K) = + 51.3 KJ; delta-1 H degree (CF_2, 298 K) = + 33.2 kJ The vapour pressure of a gas X_2 is 16 kPa (0.16 bar) at 298 K, and 100 kPa at 350 K. Find What additional information would you need to find delta C_p ? What can you say about.

Explanation / Answer

3a) Given rG0 (NO2, 298 K) = +51.3 Kj; we employ the relation

rG0 = -R*T*ln Keq where Keq = equilibrium constant for the reaction

N2 + 2 O2 -------> 2 NO2

Plug in the values ( T = 298 K) to obtain Keq as

(+51.3 kJ) = -(8.314 J/mol.K)*(298 K)*ln Keq

====> +51.3*103 J = -(2477.572 J/mol)*ln Keq

====> ln Keq = (+51.3*103)(-2477.572) = -20.7057 (I have kept a few guard digits extra).

====> Keq = e-20.7057 = 1.0177*10-9 (ignore units here since Keq should be dimensionless).

The equilibrium constant can be written as

Keq = [NO2]2/[N2][O2]2

Set up the ICE chart to find out the equilibrium constants:

N2 + 2 O2 <=====> 2 NO2

Initial                                                          1        1                          0

Change                                                     -x      -2x                      +2x

Equilibrium                                          (1 – x) (1 – 2x)                  2x

Plug in the equilibrium constant expression:

Keq = [NO2]2/[N2][O2]2 = (2x)2/(1-x).(1-2x)2

=====> 1.0177*10-9 = 4x2/(1-x)(1-2x)2

We see that Keq is of the order of 10-9 which essentially means that Keq is quite small; N2 and O2 doesn’t react appreciably at 298 K and hence, we should practically have all of N2 and O2 left in the reaction mixture with insignificant concentration of NO2. Therefore, we can assume (1 – x) 1 and (1 – 2x) 1; therefore, plug in the above expression to write

Keq = 4x2/(1).(1) = 4x2

====> 1.0177*10-9 = 4x2

====> x2 = 2.54425*10-10

====> x = 1.595*10-5

Therefore, the equilibrium concentration of NO2 = 1.595*10-5 and the equilibrium concentration of N2 = (1 – 1.595*10-5) mole = 0.99998 1.0

Therefore, [N2]/[NO2] = (1 mole)/(1.595*10-5) = 62695.9247 = 6.269*104 (ans).

b) We need to calculate rG at 350 K. I believe the question wanted you to put fH0 for NO2 at 298 K (I am not sure why ClF3 is given when it doesn’t occur in the equation. Is this fH0 for NO2? We need some information about fH0 to calculate rG at 350 K.

rG0 = fH0 – T*fS0

=====> (+51.3 kJ) = (+33.2 kJ) – (298 K)*fS0

=====> fS0 = (51.3 – 33.2 kJ)/(298 K) = 60.738 JK-1

We need to calculate rG at 350 K by noting that fH0 and fS0 are independent of temperature. Plug T = 350 K in expression below and obtain

rG = fH0 – T*FS0 = (+33.2 kJ) – (350 K)*(60.738 JK-1) = 11.9417 kJ

Use the equation

rG = rG0 + RTlnQ where Q is the reaction quotient. Plug in the values to obtain

(+11.9417 kJ) = (+51.3 kJ) + (8.314 J/mol.K)*(350 K)*ln Q

====> ln Q = -13.5256

====> Q = e-13.5256 = 1.336*10-6

As before, the reaction quotient can be written as

Q = [NO2]2/[N2][O2]2 = (2x’)2/(1-x’)(1-2x’)2

Assume x’<< 1 and write,

1.336*10-6 = 4x’2/(1).(1)

====> x’ = 5.779*10-4

Therefore, at 350 K, [NO2] = 5.779*10-4 mole and [N2] = 1 mole; therefore, [N2]/[NO2] = (1 mole)/(5.779*10-4 mole) = 1.730*10-3 (ans).

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