Consider the following equilibrium: Now suppose a reaction vessel is filled with

Question

Consider the following equilibrium: Now suppose a reaction vessel is filled with 7.56 atm of dinitrogen tetroxide (N20 at 646. °C. Answer the following questions about this system: O rise Under these conditions, will the pressure of N2O4 tend to rise or fall? x10 O fall Is it possible to reverse this tendency by adding NO2? In other words, if you said the pressure of N2O,wl ted to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N204 will tend to fall, can that be changed to a tendency to rise by adding NO2 If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO2 needed to reverse it. Round your answer to 2 significant digits. O yes O no atm

Explanation / Answer

N2O4 (g) <-----> 2NO2 ; dG = 5.4 kJ

dG = - 2.303RTlogK

5.4 * 10^3 = -2.303 * 8.314 * (646+273)*logK

K = 0.4933

                N2O4 (g) <-----> 2NO2 ; Kp = 0.4933

initially    7.56              0

at equili (7.56-x)          x

Kp = (p NO2)^2/P N2O4

0.4933 = x^2/(7.56-x)

x = 1.7 atm = equilibrium preesure of NO2

Equilibrium pressure of N2O4 = 7.56 - 1.7 = 5.86 atm

The pressure of N2O4 fall.

If you add NO2 then the number of moles of NOe increases sothat the presuure on product side increases and the reaction will shift in backward reaction.

If you increase the pressure of NO2 by greater than 5.86 atm then the reaction will reversed.

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