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 is: K_c = ([Cl₂]^2/[SO₂Cl₂]) Substitute the given values and let the equilibrium concentration of Cl₂ be x: 0.045 = (x^2/(0.075-x)) Solving for x, we get: x = 0.0398 M Since we started with 0.075 M and used up some to form x, the equilibrium concentration of SO₂Cl₂ is (0.075 - x) = 0.0352 M. Therefore, the equilibrium concentration of Cl₂ is 2x = 0.0796 M. However, we need to take into account the stoichiometry of the reaction, which tells us that the equilibrium concentration of Cl₂ is equal to the equilibrium concentration of SO₂. Therefore, the correct answer is 0.040 M.

Sulfuryl Chloride Decomposes to Sulfur Dioxide and Chlorine