Consider the following equilibrium 2NH3(g)-N2(g) + 3H2(g) AGo = 34. kJ Now suppo

Question

Consider the following equilibrium 2NH3(g)-N2(g) + 3H2(g) AGo = 34. kJ Now suppose a reaction vessel is filled with 7.15 atm of ammonia (NH) and 2.91 atm of hydrogen (H2) at 1136. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NH3 tend to rise or fall? fall Is it possible to reverse this tendency by adding N2 In other words, if you said the pressure of NH3 will tend to rise, can that be changed to a tendency to fall by adding N2? Similarly, if you said the pressure by adding N2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of N2 needed to reverse it. Round your answer to 2 significant digits yes no of NH3 will tend to fall, can that be changed to a tendency to rise atm

Explanation / Answer

For the given equilibrium reaction,

2NH3(g) <==> N2(g) + 3H2(g)    dGo = 34 kJ

When reaction vessel is filled with 7.15 atm of NH3 gas and 2.91 atm of H2 has, at 1136 oC, then,

Under these conditions, will the pressure of NH3 rise of fall : Fall

Is it possible to reverse this trndency by adding N2 : Yes

In other words, if you said the pressure of NH3 will tend to rise, can that be changed to a tendency to fall by adding N2. Similarly, if you said the pressure of NH3 will tend to fall, can that be changed to a tendency to rise by adding N2 : Yes

If you said the tendency can be reversed in the second question, calculation for the minimum pressure of N2 needed to reverse it.

Using,

dG = dGo + RTlnQ

with,

dG = 0 when reaction is reversed

dGo = 34 kJ

Q = (p[N2].p[H2]^3/p[NH3]^2)

R = gas constant

T = 1136 oC + 273 = 1409 K

So,

dGo = -RTlnQ

34 = -(0.0083145)(1409)ln(p[N2] x (2.91)^3/(7.15)^2)

p[N2] = 0.11 atm

So the pressure of N2 needed to reverse the reaction is 0.11 atm

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