Question

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)

Answer 1

Answer:

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.  

$\text{calculating the initial HI}= \frac{mol}{V}$

                                       $=\frac{9.3 \times 10 ^ -3}{2}$

                                      $=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

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:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2$

$HI \ eq= 0.00465 - 2x$

          $=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:  

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$, you can calculate the value of Kc at 1000 K.  

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

         $= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

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

$Kc = \frac{1}{Kc}$

     $= \frac{1}{0.034386}\\ \\= 29.081$

Answer 2

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

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