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
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]
