Hydrogen peroxide may decompose to form water and oxygen gas according to the following reaction. 2H₂O₂(g) 2H₂O(g) + O₂(g)
In a particular experiment,1.75 moles of H₂O₂ were placed in a 2.5-L reaction chamber at
307ºC.After equilibrium was reached,1.20 moles of H₂O₂ remained.What is K_c for the reaction?

Question

Hydrogen peroxide may decompose to form water and oxygen gas according to the following reaction. 2H₂O₂(g)

Hydrogen peroxide may decompose to form water and oxygen gas according to the following reaction. 2H<sub>2</sub>O<sub>2</sub>(g) 2H<sub>2</sub>O(g) + O<sub>2</sub>(g) In a particular experiment,1.75 moles of H<sub>2</sub>O<sub>2</sub> were placed in a 2.5-L reaction chamber at 307ºC.After equilibrium was reached,1.20 moles of H<sub>2</sub>O<sub>2</sub> remained.What is K<sub>c</sub> for the reaction? A) 2.0 × 10<sup>-4</sup> B) 2.3 × 10<sup>-2</sup> C) 2.4 × 10<sup>-3</sup> D) 5.5 × 10<sup>-3</sup> E) 3.9 × 10<sup>-4</sup>

2H₂O(g) + O₂(g)
In a particular experiment,1.75 moles of H₂O₂ were placed in a 2.5-L reaction chamber at
307ºC.After equilibrium was reached,1.20 moles of H₂O₂ remained.What is K_c for the reaction?

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: Kc = [H₂O]²/[H₂O₂] At equilibrium, the concentration of H₂O₂ is 1.20 moles/2.5 L = 0.48 M. The concentration of H₂O can be calculated as follows: Initial concentration = 1.75 moles/2.5 L = 0.70 M Equilibrium concentration = Initial concentration - x (where x is the amount of H₂O that decomposes) = 0.70 M - 0.48 M = 0.22 M Substituting these values in the equilibrium constant expression: Kc = (0.22)²/(0.48) = 0.1004 M⁻¹ Using scientific notation, this can be written as 2.3 × 10⁻². Therefore, the correct answer is B.

Hydrogen Peroxide May Decompose to Form Water and Oxygen Gas