In the reaction PCl5(g)leftharpoons PCl3(g)+Cl2(g). the equilibrium concentrations of textPC textl text5 and textPC textl text3 are 0.4 and 0.2 mol/L respectively. If the value of textKc is 0.5, what is the concentration of textC textl text2 in mol/L?

Question

In the reaction PCl5(g)leftharpoons PCl3(g)+Cl2(g). the equilibrium concentrations of textPC textl text5 and textPC textl text3 are 0.4 and 0.2 mol/L respectively. If the value of textKc is 0.5, what is the concentration of textC textl text2 in mol/L? (A) 2.0 (B) 1.5 (C) 1.0 (D) 0.5

Tardigrade Admin · Accepted Answer

underset beginsmallmatrix text0 texttime text1 textmole textatequili 1-x textbrium =0.4 endsmallmatrix mathopPCl5 leftharpoons underset beginsmallmatrix 0 mole x 0.2 endsmallmatrix mathopPCl3 + underset beginsmallmatrix 0 mole x - endsmallmatrix mathopCl2 Kc=([PCl3][Cl2]/[PCl5])Kc=0.5 0.5=(0.2× [Cl2]/0.4) [Cl2]=(0.4× 0.5/0.2)= text1 textmol/L

Q. In the reaction $PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g) .$ the equilibrium concentrations of $PC l_{5}$ and $PC l_{3}$ are 0.4 and 0.2 mol/L respectively. If the value of $K_{c}$ is 0.5, what is the concentration of $C l_{2}$ in mol/L?

2.0

1.5

1.0

0.5

$0 time 1 mole atequili 1 - x brium = 0.4 PC l_{5} ⇌ 0 m o l e x 0.2 PC l_{3} + 0 m o l e x - C l_{2}$ $K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]} K_{c} = 0.5$ $0.5 = \frac{0.2 \times [ C l _{2} ]}{0.4}$ $[C l_{2}] = \frac{0.4 \times 0.5}{0.2} = 1 mol/L$

Q. In the reaction PCl5​(g)⇌PCl3​(g)+Cl2​(g). the equilibrium concentrations of PCl5​ and PCl3​ are 0.4 and 0.2 mol/L respectively. If the value of Kc​ is 0.5, what is the concentration of Cl2​ in mol/L?

2.0

1.5

1.0

0.5

0time1moleatequili1−xbrium=0.4​PCl5​​⇌0molex0.2​PCl3​​+0molex−​Cl2​​ Kc​=[PCl5​][PCl3​][Cl2​]​Kc​=0.5 0.5=0.40.2×[Cl2​]​ [Cl2​]=0.20.4×0.5​=1mol/L

Q. In the reaction $PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g) .$ the equilibrium concentrations of $PC l_{5}$ and $PC l_{3}$ are 0.4 and 0.2 mol/L respectively. If the value of $K_{c}$ is 0.5, what is the concentration of $C l_{2}$ in mol/L?

$0 time 1 mole atequili 1 - x brium = 0.4 PC l_{5} ⇌ 0 m o l e x 0.2 PC l_{3} + 0 m o l e x - C l_{2}$ $K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]} K_{c} = 0.5$ $0.5 = \frac{0.2 \times [ C l _{2} ]}{0.4}$ $[C l_{2}] = \frac{0.4 \times 0.5}{0.2} = 1 mol/L$