Initially, 0.8 mol of PCl5 and 0.2 mol of PCl3 are mixed in a one litre vessel. At equilibrium, 0.4 mol of PCl3 is present. The value of Kc for the reaction, PCl5(g)leftharpoons PCl3(g)+Cl2(g), would be

Question

Initially, 0.8 mol of PCl5 and 0.2 mol of PCl3 are mixed in a one litre vessel. At equilibrium, 0.4 mol of PCl3 is present. The value of Kc for the reaction, PCl5(g)leftharpoons PCl3(g)+Cl2(g), would be (A) 0.13 text mol textL- 1 (B) 0.66 text mol textL- 1 (C) 0.013 text mol textL- 1 (D) 0.05 text mol textL- 1

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/c-zmxcvjjth5o1.png" /> So x=0.2 Kc=([P C l3] [C l2]/[P C l5])=(0.4 × 0.2/0.6) =0.13 text mol textL- 1

Q. Initially, 0.8 mol of $PC l_{5}$ and 0.2 mol of $PC l_{3}$ are mixed in a one litre vessel. At equilibrium, 0.4 mol of $PC l_{3}$ is present. The value of $K_{c}$ for the reaction, $PC l_{5} (g) ⇌ PC l_{3} (g) + C l_{2} (g),$ would be

$0.13 mol L^{- 1}$

$0.66 mol L^{- 1}$

$0.013 mol L^{- 1}$

$0.05 mol L^{- 1}$

So $x = 0.2$

$K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]} = \frac{0.4 \times 0.2}{0.6}$

$= 0.13 mol L^{- 1}$

Q. Initially, 0.8 mol of PCl5​ and 0.2 mol of PCl3​ are mixed in a one litre vessel. At equilibrium, 0.4 mol of PCl3​ is present. The value of Kc​ for the reaction, PCl5​(g)⇌PCl3​(g)+Cl2​(g), would be

0.13 mol L−1

0.66 mol L−1

0.013 mol L−1

0.05 mol L−1