1. Tardigrade
  2. Question
  3. Chemistry
  4. 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 Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

EAMCETEAMCET 2003

Solution:

$ \underset{\begin{smallmatrix} \text{0}\,\text{time}\,\text{1}\,\text{mole} \\ \text{atequili}\,1-x\, \\ \text{brium}\,=0.4 \end{smallmatrix}}{\mathop{PC{{l}_{5}}}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0\,mole \\ \,\,\,\,x \\ \,\,\,0.2 \end{smallmatrix}}{\mathop{PC{{l}_{3}}}}\,+\underset{\begin{smallmatrix} 0\,mole \\ \,\,\,x \\ \,\,\,\,- \end{smallmatrix}}{\mathop{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}=\text{1}\,\text{mol/L} $