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Consider the following equilibrium N_2O_4 (g) doubleheadarrow 2NO_2 (g) Thermody

ID: 525614 • Letter: C

Question

Consider the following equilibrium N_2O_4 (g) doubleheadarrow 2NO_2 (g) Thermodynamic data on these gases can be found in Appendix or equivalent tables. You may assume that Delta H degree & Delta S degree do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much NO_2 as N_2O_4? (c) At what temperature will an equilibrium mixture of 10 atm total pressure contains twice as much NO_2 as N_2O_4? (d) Rationalize the results from parts (b) and (c) by using lessthanorequalto Chatelier's Principle.

Explanation / Answer

Let's consider 1 mole of N2O4(g).

The balanced chemical reaction is N2O4(g)2NO2(g)

Enthalpy change for the above reaction is

DeltaH0 = 2DeltaHf0(NO2) - DeltaHf0(N2O4) = 2x 33.18 KJ/mol - 9.16 KJ/mol =58.02 kj

H = [2(33.84 kJ)] – [9.66 kJ] = 58.02 k

JS = [2(240.45 J)] – [304.3 J] = 176.6 J

Go= 58.02 kJ – T(0.1766 kJ/K)

At equilibrium,G = 0;G =Go+ RTlnK

0 =Go+ RTln[(PNO2)2/PN2O4]Æ0 =Gowhen PNO2= PN2O4= 1 atm

0 = 58.02 kJ – T(0.1767 kJ/K);

T= 58.02/0.1767 = 328.5 K------------answer

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At what temperature in Kelvin will an equilibrium mixture of 1 atm total pressurecontain twice as much NO2as N2O4?

Ptot= PNO2+ PN2O4Æ1 atm = 2x + xÆx = 1/3 atm

0 =Go+ RTln[(PNO2)2/PN2O4]

0 = 58.02 kJ – T(0.1766 kJ/K) + RTln[(2/3)2/(1/3)]

0 = 58.02 kJ – T(0.1766 kJ/K) + RT(0.2877)

58.02 kJ = 0.1743 T;T = 333 K--------answer

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(c)At what temperature in Kelvin will an equilibrium mixture of 10 atm total pressurecontain twice as much NO2as N2O4?374.2

As above, but now x = 10/3 atm;

0 = 58.02 kJ – T(0.1551 kJ/K); T = 374.2 K

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(d)Rationalize the results from parts(b)and(c)

by using Le Châtelier’s principle

The reaction is endothermic, so increasing the amount of product that'spresent requires a temperature increase...induce a right-shift--------answer