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 3.36 atm of nitrogen (N 2) and 8.32 atm of ammonia (NH3) at 564·°C Answer the following questions about this system rise Under these conditions, will the pressure of NH3 tend to rise or fall? fall ts it possible to reverse this tendency by adding H2? In other words, if you said the pressure of NH2 will tend to rise, can that be changed to a tendency to fall by adding H Similarly, if you said the pressure of NH2 will tend to fall, can that be changed to a tendency to rise by adding H,? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H2 needed to reverse it Round your answer to 2 significant digits. yes O no atm

Explanation / Answer

a)

since we do NOT have H2 gas, then th ereaciton must favour H2 production, so the lef tside must increase

then

NH3 must fall/decrease to favour H2

b)

yes we can, by adding H2

c)

Q = K

dG = -34 --> -34000

dG = -rT*ln(K)

K = exp(-dG/(RT)) = exp(34000/(8.314*(564+273)))

k = 132.407

K = [NH3]^2 / [N2][H2]^3

132.407 = (8.32^2) /(3.36) / (H2^3)

H2^3 =  (8.32^2) /(3.36) /132.407

H2 = (0.155)^(1/3)

P-H2 = 0.5371 atm

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