Question

ADVANCED MATERIAL Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NH3 (8) -N,()+3H, 68) AG" = 34. kJ Now suppose a reaction vessel is filled with 5.74 atm of ammonia (NH) and 1.66 atm of nitrogen (N2) at 220. °C. Answer the following questions about this system: Under these conditions, will the pressure of N, tend to rise or fall? x 5 ? yes Is it possible to reverse this tendency by adding H2? In other words, if you said the pressure of N, will tend to rise, can that be changed to a tendency to fall by adding H,? Similarly, if you said the pressure of N, will tend to fall, can that be changed to a tendency to rise by adding H2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H, needed to reverse it. Round your answer to 2 significant digits.

Answer 1

Under these conditions, will the pressure of N₂ tend to rise or fall? : rise

Is it possible to reverse this tendency by adding H₂? : yes

minimum pressure of H₂ = 0.17 atm

Explanation

$\Delta$G^o = 34. kJ

ln(K_p) = -($\Delta$G^o) / (R * T)

where K_p is pressure equilibrium constant

R = 8.314 J/mol-K

T = temperature = 220. ^oC = 493 K

ln(K_p) = -(34 x 10³ J/mol) / [(8.314 J/mol-K) * (493 K)]

ln(K_p) = -8.3

K_p = e^-8.3

K_p = 2.5 x 10^-4

K_p = [pN₂]_eq[pH₂]_eq³ / [pNH₃]_eq²

2.5 x 10^-4 = [(1.66 atm) * [pH₂]_eq³] / (5.74 atm)²

[pH₂]_eq³ = (2.5 x 10^-4) * (5.74 atm)² / (1.66 atm)

[pH₂]_eq = (5.0 x 10^-3)^1/3

[pH₂]_eq = 0.17 atm