Question

Diatomic iodine (I2) decomposes at high temperature to form II atoms according to the reaction:

I2(g) ⇌ 2I(g), Kc=0.011 at 1200∘C
In an equilibrium mixture, the concentration of I2 is 0.21 M.
What is the equilibrium concentration of I?

Answer 1

The equilibrium concentration of I atoms when diatomic iodine decomposes at a high temperature with a given equilibrium constant and I2 concentration is calculated using the equilibrium expression, resulting in an I atom concentration of approximately 0.048 M.

Step-by-step explanation:

The student has asked about the equilibrium concentration of I atoms when diatomic iodine (I2) decomposes at high temperature. Given that the reaction is I2(g) ⇌ 2I(g), with an equilibrium constant Kc of 0.011 at 1200°C, and the concentration of I2 is 0.21 M at equilibrium, we can calculate the concentration of I using the equilibrium expression.

The equilibrium expression is Kc = [I]2/[I2], which can be rearranged to find [I] = √(Kc × [I2]). Substituting the given values, we get:

[I] = √(0.011 × 0.21 M) ≈ √(0.00231 M) ≈ 0.048 M

Therefore, the equilibrium concentration of I atoms is approximately 0.048 M.

The equilibrium concentration of I can be determined using the equilibrium constant, Kc, and the concentration of I2. In the given reaction, I2 decomposes to form 2I, so the equilibrium constant expression is [I]^2/[I2]. Given that the concentration of I2 is 0.21 M, we can substitute this value into the equilibrium constant expression to find the equilibrium concentration of I.

[I] = √(Kc × [I2]) = √(0.011 × 0.21) ≈ 0.056 M

Therefore, the equilibrium concentration of I is approximately 0.056 M.