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?

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.