Sulfuryl chloride decomposes to sulfur dioxide and chlorine. SO₂Cl₂(g) SO₂(g) + Cl₂(g)
K_c is 0.045 at 648 K.If an initial concentration of 0.075 M SO₂Cl₂ is allowed to equilibrate,what is the equilibrium concentration of Cl₂?

Question

Sulfuryl chloride decomposes to sulfur dioxide and chlorine. SO₂Cl₂(g)

Sulfuryl chloride decomposes to sulfur dioxide and chlorine. SO<sub>2</sub>Cl<sub>2</sub>(g) SO<sub>2</sub>(g) + Cl<sub>2</sub>(g) K<sub>c</sub> is 0.045 at 648 K.If an initial concentration of 0.075 M SO<sub>2</sub>Cl<sub>2</sub> is allowed to equilibrate,what is the equilibrium concentration of Cl<sub>2</sub>? A) 0.0034 M B) 0.030 M C) 0.040 M D) 0.058 M E) 0.075 M

SO₂(g) + Cl₂(g)
K_c is 0.045 at 648 K.If an initial concentration of 0.075 M SO₂Cl₂ is allowed to equilibrate,what is the equilibrium concentration of Cl₂?

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is: K_c = [Cl₂] / [SO₂Cl₂][Cl₂] At equilibrium, let the concentration of Cl₂ be x M, then the concentration of SO₂Cl₂ would be (0.075 - x) M. Therefore: K_c = x / (0.045 * (0.075 - x)) Solving for x gives: x = 0.040 M Therefore, the equilibrium concentration of Cl₂ is 0.040 M, which is choice C.

Sulfuryl Chloride Decomposes to Sulfur Dioxide and Chlorine