Question

Two moles of HI when heated at 700K in a sealed tube until equilibrium is reached were found to be 22% dissociated. Calculate $K_{c}$ for the dissociation.

Answer 1

The dissociation reaction of Hydrogen iodide is as follows.

$2HI\quad \rightarrow \quad \quad { H }_{ 2 }\quad +\quad \quad { I }_{ 2 }$

$\quad 1\quad \quad \quad \quad \quad \quad 0\quad \quad \quad \quad \quad 0\\ (1-x)\quad \quad \left( \frac { 1 }{ x }  \right) \quad \quad \left( \frac { 1 }{ x }  \right)$

From the given,

22% of HI is dissociated.  

$For\quad remaining\quad =\quad (100\quad -\quad 22)%\quad =\quad 78\quad =\quad 0.78$

$For\quad one\quad mole\quad =\quad \frac { 22 }{ 100 } \quad \times \quad 1\quad =\quad 0.22$

Equilibrium constant of the dissociation reaction is as follows.

${ k }_{ c }\quad =\quad \frac { [{ H }_{ 2 }][{ I }_{ 2 }] }{ [HI{ ] }^{ 2 } }$  

Substitute the values.

$=\quad \frac { 0.11\quad \times \quad 0.11 }{ (0.78{ ) }^{ 2 } } \quad =\quad 0.0199$  

Therefore, equilibrium constant of the dissociated reaction is 0.0199

Answer 2

Hey mate.....
$K_{c}=0.0199$