Question

$PC{l}_5,PC{l}_3\text{and}C{l}_2$ are at equilibrium at 500 𝐾 and having concentration $1.59MPC{l}_3,1.59MC{l}_2\text{and}1.41MPC{l}_5$.
Calculate $K_c$ for the reaction, $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

Answer 1

Step 1: Equilibrium constant
For a general reversible reaction $aA+bB\rightleftharpoons cC+dD$, the concentrations in an equilibrium mixture are related by the following equilibrium equation.

$K_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

Step 2: Value of $K_c$
The given reaction is-
$PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$

So, the equilibrium constant $K_c$ for the above reaction can be written as,
$K_c=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}=\frac{(1.59{)}^2}{1.41}=1.79$

Final Answer:
$K_c$ for the reaction is $1.79$.