A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentration of I2 was 6.29×10^−4M. Calculate the Kc at 1000K for: H2(g)+I2(g)⇌2HI(g)

The Kc of the reaction is 40.

Molarity of the HI = 9.30×10^−3mol/ 2.00 L = 4.65 × 10^-3 M

Let the concentrations of I2 and H2 be x, but we are told in the question that 6.29×10^−4M was present at equilibrium.

The molarity of HI at equilibrium now becomes; 4.65 × 10^-3 M - 6.29×10^−4M

= 4 × 10^-3 M

But;

Kc = [HI]^2/[H2] [I2]

Kc = ( 4 × 10^-3)^2/(6.29×10^−4)^2

Kc = 40

Learn more: link

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

It is defined that:

Now, we calculate the position:  

For the reaction , you can calculate the value of Kc at 1000 K.  

data expression for Kc

calculating the reverse reaction

its mg hope this

i just learnt this recently but i need to look through my notes again