The reaction #"HCl(c) + NH"_3(g) ⇌ "NH"_4"Cl(s)"# is spontaneous only at low temperatures. What are the answers to the following questions?

(a) What is the sign of #ΔS# for the forward direction?
(b) What is the value of #ΔG# at equilibrium?
(c) What happens to the sign of #ΔG# as the temperature increases?
(d) What is the temperature in terms of #ΔH#and #ΔS# at the point where #ΔG# changes sign?
(e) How does the addition of more ammonium chloride affect the value of the equilibrium constant?
(f) How does an increase in temperature affect the value of the equilibrium constant?

1 Answer

Answer:

Here's what I get.

Explanation:

HCl(g) + NH₃(g) ⇌ NH₄Cl(s)

(a) The sign of #ΔS# for the forward reaction is negative.

You are converting 2 mol of gas to a solid. The freedom of motion of the gas particles decreases, so the entropy decreases.

(b) #ΔG = 0# at equilibrium.

#ΔG = 0# for any reaction that is at equilibrium.

(c) The forward reaction is spontaneous at low temperatures. When the temperature of the reaction increases, the sign of #ΔG# changes.

That means that #ΔG# is negative: #ΔG = 0#.

#ΔH –TΔS < 0#

#ΔH# is probably negative.

But #ΔS# is negative, so #TΔS# is negative, and #-TΔS# is positive.

If #T # is large enough, the #–TΔS# term will become so large and positive that it overcomes the #ΔH# term, and #ΔG# will become positive.

(d) At the point when the sign changes, #ΔG = 0#. At that point,

#T = (ΔH)/(ΔS)#

(e) The value of #K# is not affected by adding additional solid NH₄Cl.

The only thing that changes the value of #K# is changing the temperature.

(f) The value of #K# will decrease as the temperature increases.

We saw in Part (c) that #ΔG# becomes positive when the temperature is high enough.

The reaction becomes spontaneous in the reverse direction, so #K# must decrease as the temperature increases.