Question

At equilibrium, K for the decomposition of HI(g) was found to be 1.07 x 10^-5. The equilibrium concentration of HI(g) was found to be 0.129M. Calculate the concentration of I2 at equilibrium. (Hint – Let x = the concentration of I2. What would the concentration of H2 be if x is the concentration of I2? Refer to the coefficients of the equation to help you.) 2HI(g) → H2(g) + I2(g)

a) 6.45 x 10^-6 M
b) 8.55 x 10^-7 M
c) 1.07 x 10^-5 M
d) 0.129 M

Answer 1

To calculate the concentration of I2 at equilibrium in the decomposition of HI, we can use the given equation and equilibrium constant. By setting x as the concentration of I2 and using the stoichiometric coefficient, we can solve for x using the equilibrium constant expression. The concentration of I2 at equilibrium is found to be 6.45 x 10^-6 M.

Step-by-step explanation:

To determine the concentration of I2 at equilibrium, we can use the equation provided and the information given. Let x be the concentration of I2. Since the equation has a stoichiometric coefficient of 1 for I2, the concentration of H2 would also be x. The equilibrium concentration of HI is given as 0.129 M.

Using the equilibrium constant expression, K = [H2][I2]/[HI]^2, we can substitute the given values and solve for x:

K = (x)(x)/(0.129)^2

Simplifying the equation gives x^2 = (K)(0.129)^2

Taking the square root of both sides gives x = sqrt[(K)(0.129)^2]

Substituting the given value of K = 1.07 x 10^-5, we can calculate x:

x = sqrt[(1.07 x 10^-5)(0.129)^2] = 6.45 x 10^-6 M

Therefore, the concentration of I2 at equilibrium is 6.45 x 10^-6 M.