Question

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

Answer 1

Answer:

Equilibrium concentration of reactants and products:

$[HI]=(0.372-2x) M =(0.372-2\times 0.03935)M =0.2933 M$

$[H_2]=x = 0.03935 M$

$[I_2]=x = 0.03935 M$

Explanation:

The equilibrium constant of the reaction = $K_c=1.80\times 10^{-2}$

Mole sof HI = 0.372 mol

Volume of the vessel = 1.00 L

Initial concentration of HI = $\frac{0.372 mol}{1.00 L}=0.372 M$

$2HI(g)\rightarrow H_2(g) + I_2(g)$

0.372 M

At equilibrium

(0.372-2x) M          x               x

An expression of an equilibrium constant will be given as;

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

$1.80\times 10^{-2}=\frac{x\times x}{(0.372-2x)^2}$

Solving for x;

x = 0.03935

Equilibrium concentration of reactants and products:

$[HI]=(0.372-2x) M =(0.372-2\times 0.03935)M =0.2933 M$

$[H_2]=x = 0.03935 M$

$[I_2]=x = 0.03935 M$