Question

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.

Answer 1

`K_c = 51`

Step-by-step 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

`K_c = ([HI]^2)/([H_2][I_2]) \\\\= ((1.87*10^-^3)^2)/((6.5*10^-^5)(1.065*10^-3)) \\\\K_c = 51`

Answer 2

1.74845

Step-by-step 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