Question

For the reaction I2(g) ⇌ 2 I(g), the value of Kc = 3.76 x 10^−5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]?

A. [I2] = 1.88 x 10^−5 M, [I] = 3.76 x 10^−5 M
B. [I2] = 3.76 x 10^−5 M, [I] = 1.88 x 10^−5 M
C. [I2] = 1.00 M, [I] = 2.00 M
D. [I2] = 2.00 M, [I] = 1.00 M

Answer 1

Using an ICE table and the equilibrium constant Kc, we set up a quadratic equation to solve for 'x', representing the change in concentration for I2 and I, and calculate their equilibrium concentrations in the context of the given reaction at 1000 K.

Step-by-step explanation:

For the equilibrium process I2(g) ⇌ 2 I(g), where the equilibrium constant (Kc) is 3.76 x 10⁻⁵ at 1000 K, we begin with 1.00 mole of I2 in a 2.00 L flask.

To find the equilibrium concentrations, we can set up an ICE table (Initial, Change, Equilibrium). Initially, there is 1.00 mole of I2, which gives an initial concentration of 1.00 mole / 2.00 L = 0.500 M of I2. No I atoms are present initially. At equilibrium, the concentration of I2 will have decreased by 'x' and the concentration of I will have increased by '2x' (since the reaction produces two moles of I for each mole of I2 that reacts).

The equilibrium concentrations can be represented as [I2] = 0.500 - x and [I] = 2x. Using the equilibrium expression Kc = [I]2 / [I2], we substitute the known value of Kc and expressions for [I2] and [I] to solve for 'x'. This quadratic equation can then be solved to find the value of x, and subsequently the equilibrium concentrations of I2 and I.