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
Answer:
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
