Question

3.2 moles of HI(g) was heated in a sealed bulb at ${444}^{0}C$ till the equilibrium was reached. Its degree of dissociation was found to be $20\%$.Calculate the number of moles of hydrogen iodide, hydrogen, and iodine present at the equilibrium point and determine the value of the equilibrium constant for the reaction $2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$. Considering the volume of the container 1 L.

Answer 1

$2{HI}_{\left(g\right)}\rightleftharpoons {H_2}_{\left(g\right)}+{I_2}_{\left(g\right)}$

Given that:-

Degree of dissociation $\left(\alpha \right)=0.2$

Initially-

No. of mles of $HI=3.2\text{mole}$

Therefore, at equillibrium-

No. of moles of $HI=3.2(1-\alpha )=3.2(1-0.2)=2.56\text{mole}$

No. of moles of $HI$ reacted $=3.2-2.56=0.64$

According to reaction-

$2$ moles of $HI$ gives $1$ mole of $H_2$ and $I_2$ each.

$\therefore 0.64$ moles of $HI$ will give $0.32$ mole of $H_2$ and $I_2$ each.

No. of moles of $H_2$ at equillibrium $=0.32\text{mole}$

No. of moles of $I_2$ at equillibrium $=0.32\text{mole}$

For the above reaction-

$K_c=\frac{\left[H_2\right]\left[I_2\right]}{{\left[HI\right]}^2}=\frac{0.32\times 0.32}{{\left(2.56\right)}^2}=0.0156$