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.

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; /// i do believe that the correct answer is (c)nuclear bond