Consider uie following equilibrium Now suppose a reaction vessel is filled with

Question

Consider uie following equilibrium Now suppose a reaction vessel is filled with 629 atm of dinitrogen tetroxide (N204) at 388· Answer the following questions about this system: rise Under these conditions, will the pressure of N204 tend to rise or fall? fall Is it possible to reverse this tendency by adding NO2? In other words, if you said the pressure of N204 will tend to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N20, 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 no 1Mm

Explanation / Answer

For the reaction,

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

When the system has 6.29 atm of N2O4(g), at 388 oC then,

Under these conditions, the pressure of N2O4(g) will rise or fall : Fall

Is it possible to reverse this tendency by adding NO2 : Yes

In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO. Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to rise by adding NO2 : Yes

If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO2 needed to reverse it.

Using,

dG = dGo + RTlnQ

with,

dG = 0 when the reaction is reversed

therefore,

dGo = -RTlnQ

Q = (p[N2O4]/p[NO2]^2)

p = pressure

For the reaction,

R = 0.0083145 kJ/mol.K

T = 388 oC + 273 = 661 K

dGo = -5.4 kJ

p[N2O4] = 6.29 atm

So,

-5.4 = -(0.0083145)(661)ln(6.29/p[NO2]^2)

p[NO2] = 1.53 atm

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