at 603K, Kc for the reaction NH4Cl(g) ==> NH3(g) + HCl(g) is 1.1X10^(-2). what is the equilibrium concentration of NH3(g) if 6the initial concentration of NH4Cl(g) is 0.500M

Question



Answer 1

Set up an ICE table ...

NH₄Cl(g) ==> NH₃(g) + HCl(g)

0.5...................0..............0..........Initial

-x..................+x..............+x.........Change

0.5-x...............x...............x..........Equilibrium

The expression for Kc is

Kc = [NH₃][HCl] / [NH₄Cl]

From the ICE table, we can obtain these values

1.1x10^-2 = (x)(x) / (0.5-x)

Solve for x and that will give you the [NH₃] at equilibrium.

Answer 2

If x is the concentration of NH₄Cl gas reacted in order to reach equilibrium, then the concentrations of NH₄Cl, NH₃, and HCl at equilibrium are .5 -x, x, and x. You can determine this using an ICE diagram.

K_C expression is C_NH3C_HCl/C_NH4Cl

^{at equilibrium K}_C = 1.1 x 10^-2 = x²/(.5-x)

Multiply out and solve the quadratic equation x² + .011x -.0055 = 0 using quadratic formula. Only one of the solutions will make physical sense.