The equilibrium constant, Kc, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) Calculate the equilibrium concentrations of reactant and products when 0.372 moles of HI are introduced into a 1.00 L vessel at 698 K. [HI] = M [H2] = M [I2] = M

Equilibrium concentration of reactants and products:

Explanation:

The equilibrium constant of the reaction =

Mole sof HI = 0.372 mol

Volume of the vessel = 1.00 L

Initial concentration of HI =

0.372 M

At equilibrium

(0.372-2x) M          x               x

An expression of an equilibrium constant will be given as;

Solving for x;

x = 0.03935

Equilibrium concentration of reactants and products:

10

Explanation:

Equation: d = m/v

100/10 = 10