NH4I(s) NH3(g) HI(g) Calculate the equilibrium concentration of HI when 0.343 moles of NH4I(s) are introduced into a 1.00 L vessel at 673 K. [HI]

This question is incomplete, the complete question is;

The equilibrium constant, Kc, for the following reaction is 7.00×10-5 at 673 K. NH4I(s) ---> NH3(g) + HI(g) Calculate the equilibrium concentration of HI when 0.343 moles of NH4I(s) are introduced into a 1.00 L vessel at 673 K

the equilibrium concentration of HI is 1.302 × 10⁻⁵ M

Explanation:

Given that;

kc = 7.00 × 10⁻⁵ at 673 K

equilibrium concentration of HI = ?

when 0.343 moles of NH4I(s) are introduced into a 1.00 L vessel at 673 K

Molarity of NH4I = 0.343 / 1.00 = 0.343 M

NH4I(s) > NH3(g) + HI(g)

0.343                             0           0   > Initial

(0.343 - x)                        x            x   > equilibrium

we know that; solid does not take part in equilibrium constant expression

so

KC = [NH3] [HI] / NH4I

we substitute

7.00 × 10⁻⁵ = x² / (0.343 - x)  

0.00007 (0.343 - x) = x²

0.00002401 - 0.00007x = x²

x² + 0.00007x - 0.00002401 = 0

using; x = -b±√(b² - 4ac) / 2a

we substitute

x = -0.00007 ± √((0.00007)² - 4×1×-0.00002401) / 2×1

x = [-0.00007 ± √( 4.9×10⁻⁹ + 9.604×10⁻⁵)] / 2

x = [-0.00007 ± 0.00009604] / 2

Acceptable value of x = [-0.00007 + 0.00009604] / 2

x = 0.00002604 /2

x = 0.00001302 ≈ 1.302 × 10⁻⁵ M

therefore the equilibrium concentration of HI is 1.302 × 10⁻⁵ M