The K c for the reaction,
PCl₅ (g) PCl₃(g) + Cl₂ (g) , at a particular temperature is 4.0*10^{- 2} .
How many moles of PCl₅ must be placed in an empty 1.0 liter flask to have the equilibrium concentration of Cl₂ = 0.15 M ?

Question

The K c for the reaction,
PCl₅ (g) The K c for the reaction, PCl5 (g) PCl3(g) + Cl₂ (g) , at a particular temperature is 4.0*10- 2 . How many moles of PCl5 must be placed in an empty 1.0 liter flask to have the equilibrium concentration of Cl₂ = 0.15 M ? A) 0.71 B) 0.15 C) 1.4 D) 6.7 E) 0.23

The K c for the reaction, PCl<sub>5</sub> (g) PCl<sub>3</sub>(g) + Cl₂ (g) , at a particular temperature is 4.0*10<sup>- 2</sup> . How many moles of PCl<sub>5</sub> must be placed in an empty 1.0 liter flask to have the equilibrium concentration of Cl₂ = 0.15 M ? A) 0.71 B) 0.15 C) 1.4 D) 6.7 E) 0.23

PCl₃(g) + Cl₂ (g) , at a particular temperature is 4.0*10^{- 2} .
How many moles of PCl₅ must be placed in an empty 1.0 liter flask to have the equilibrium concentration of Cl₂ = 0.15 M ?

Quizplus · Accepted Answer

The equilibrium expression is: K c = [PCl₃][Cl₂]/[PCl₅] Let x be the number of moles of PCl₅ that must be placed in the flask to have the equilibrium concentration of Cl₂ = 0.15 M. Therefore, [Cl₂] = 0.15 M [PCl₅] = x/1.0 L = x M (since the flask is empty) [PCl₃] = 0 M (initially there is no product) [Cl₂] = 0 M (initially there is no product) Substitute these values in the equilibrium expression: K c = [PCl₃][Cl₂]/[PCl₅] 4.0*10^{- 2} = (0)(0)/(x) 4.0*10^{- 2} = 0 x = (0.15)(1.0)/4.0*10^{- 2} x = 0.71 mol Therefore, the answer is A: 0.71.

The K C for the Reaction