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A closed system initially containing 1×10^-3M H2 and 2×10^-3 at 448°C is allowed to reach equilibrium. Analysis of the equilibrium mixture show s that the concentration of HI is 1.87×10^ -3. Calculate Kc at 448°C for the reaction.

Respuesta :

jmonterrozar

Answer:

1.74845

Explanation:

We have the following reaction:

I2 + H2 => 2 HI

Now, the constant Kc, has the following formula:

Kc = [C] ^ c * [D] ^ d / [A] ^ a * [B] ^ b

In this case I2 is A, H2 is B and C is HI

We know that the values are:

 H2 = 1 × 10 ^ -3 at 448 ° C

I2 = 2 × 10 ^ -3 at 448 ° C

HI = 1.87 × 10 ^ -3 at 448 ° C

Replacing:

Kc = [1.87 × 10 ^ -3] ^ 2 / {[2 × 10 ^ -3] ^ 1 * [1 × 10 ^ -3] ^ 1}

Kc = 1.87 ^ 2/2 * 1

Kc = 1.74845

Which means that at 448 ° C, Kc is equal to 1.74845

Olajidey

Answer:

[tex]K_c = 51[/tex]

Explanation:

[H2] = 10^-3

[I2] = 2*10^-3

[HI] = 0

in equilbiirum

[H2] = 10^-3 - x

[I2] = 2*10^-3 -x

[HI] = 0 + 2x

and we know

[HI] = 0 + 2x = 1.87*10^-3

x = ( 1.87*10^-3)/2 =  0.000935

then

[H2] = 10^-3 - 0.000935 = 0.000065

[I2] = 2*10^-3 -0.000935 = 0.001065

                        H₂                +          I              ⇄              2 HI

Initially     1 × 10⁻³                       2 × 10⁻³

Change  -9.35 × 10⁻⁴               -9.35 × 10⁻⁴                +1.87 × 10⁻³

At equil   6.5 × 10⁻⁵                  1.06 5 × 10⁻³               1.87 × 10⁻³

HI increase by 1.87 × 10⁻³M

[tex]K_c = \frac{[HI]^2}{[H_2][I_2]} \\\\= \frac{(1.87\times10^-^3)^2}{(6.5\times10^-^5)(1.065\times10^-3)} \\\\K_c = 51[/tex]

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