Consider the following equilibrium: 4NH_3(g) N_2(g) + 3H_2(g) delta G degree = 3

Question

Consider the following equilibrium: 4NH_3(g) N_2(g) + 3H_2(g) delta G degree = 34. kJ Now suppose a reaction vessel is filled with 8.92 atm of ammonia (NH_3) and 2.35 a of nitrogen (N_2) at 72. degree C. Answer the following questions about this system: Under these conditions, will the pressure of NH_3 tend to rise or fall? rise fall Is it possible to reverse this tendency by adding H_2? In other words, if you said the pressure of NH_3 will tend to rise, can that be changed to a tendency to fall by adding H_2? Similarity, if you said the pressure of NH_3 will tend to fall, can that be changed to a tendency to rise by adding H_2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H_2 needed to reverse it. Round your answer to 2 significant digits. Atm

Explanation / Answer

dG = -RT*ln(Kp)

Kp = exp(-dG/(RT))

Kp =  exp(-34000/(8.314*298)) = 0.0000010968

now, this is not favoured

if initially, we have P = 0 atm of H2, then, syustem will actually try to produce some H2

so the pressure of NH3 must decrease/fall

Q2.

the only way to avoid this is by addin H2 to the mix, so choose YES

Q3.

Q = 0.0000010968

0.0000010968= (N2)(h2)^3 /(NH3)^4

0.0000010968= (2.35)(h2)^3 /(8.92)^4

H2^3 = 0.0000010968*(8.92)^4 / 2.35

H2 = (0.00295473)^(1/3)

H2 = 0.14349 atm

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---