Question

The equilibrium constant for the reaction: $F{e}^{3+}\left(aq\right)+SC{N}^{-}\left(aq\right)\rightleftharpoons FeSC{N}^{2+}\left(aq\right)$ is 140 at 298 K.
The Equilibrium constant for the reaction: $2F{e}^{3+}\left(aq\right)+2SC{N}^{-}\left(aq\right)\rightleftharpoons 2FeSC{N}^{2+}\left(aq\right)$ is:

• A. 280
• B. 140
• C. 19600
• D. 70

Answer 1

The correct option is C 19600

$K=\frac{\left[FeSC{N}^{2+}\right]}{\left[F{e}^{3+}\right]\left[SC{N}^{-}\right]}=140,K_1=\frac{[FeSC{N}^{2+}{]}^2}{\left[F{e}^{3+}{]}^2\right[SC{N}^{-}{]}^2}=(140{)}^2=19600$