Question

Calculate $Kc$ for the reaction, $H{I}_{\left(g\right)}\to \frac{1}{2}{H}_{2\left(g\right)}+\frac{1}{2}{I}_{2\left(g\right)}$
Given that the reaction $H_2\left(g\right)+I_2\to 2HI\left(g\right)$, had been started with $1mole$ of $H_2\left(s\right)$ and $3mole$ of $I_2\left(g\right)$. At the equilibrium the moles of $HI$ formed is $1.5moles$. The volume of the reaction vessel is 2 litre.

• A. $5$
• B. $-3$
• C. $3$
• D. None
  

Answer 1

The correct option is D None

$H_2+I_2⟶2\left[HI\right]$

$K_C=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=\frac{[1.5{]}^2}{\left[1\right]\left[3\right]}=\frac{2.25}{3}=0.75$