Question

For the system $A_{\left(g\right)}+2B_{(}g)\rightleftharpoons C_{\left(g\right)}$, the equilibrium concentrations are
$A=0.06mol{L}^{-1}$
$B=0.12mol{L}^{-1}$
$C=0.216mol{L}^{-1}$
Then $K_c$ for the reaction is _____.

• A. $250$
• B. $416$
• C. $450$
• D. $525$

Answer 1

The correct option is A $250$

For Reaction $A+2B\rightleftharpoons C$ ,

$K_c=\frac{\left[C\right]}{{\left[B\right]}^2\left[A\right]}$

By Substituting the values, We get

$K_c$ = 250