Confusing Chemistry Math Question

Question

The equilibrium constant for dissociation of ammonium chloride:

NH₄Cl(s) ↔ NH₃(g) + HCl(g)

is equal to K_P = 0.013312 at 550.0 K. If 1.7 g of ammonium chloride is placed in a 1.0 L vessel and charged with 2.5 bar of HCl(g) what is the final equilibrium pressure of NH₃(g)?

Answer 1

NH₄Cl(s) ↔ NH₃(g) + HCl(g) K_P = 0.013312 at 550.0 K NH₄Cl(s) is a solid and it has no pressure

I 0 bar 0 bar 2.5 bar 1.7 g NH₄Cl(s) will decompose into NH₃(g) and HCl(g)

C +x +x

E 0 bar x 2.5 +x therefore K_p = P _NH3 P_HCl 0.013312 = x (2.5 +x)

x = -b + √b² -4ac x² +2.5 x - 0.013312 = 0 x= P _NH3 = 0.00550 bar

2a