Question

if alpha is degree of dissocliation, then the total number of moles for the reaction starting with 1 mole of HI
2HI = H2+I2 will be​

Answer 1

Explanation:

hope the answer is thankful to you....

Answer 2

Given:

Alpha is degree of dissocliation,

To find:

The total number of moles for the reaction starting with 1 mole of HI.

Calculation:

$\rm{2HI \: \longrightarrow \: H_{2} + I_{2}}$

Starting from 1 mole , reaction will be:

$\rm{HI \: \longrightarrow \: \frac{1}{2} H_{2} + \frac{1}{2} I_{2}}$

At t = 0 seconds:

$1\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0 \: \: \: \: \: \: \: \: \: \: 0$

At t = equilibrium

$(1- \alpha) \: \: \: \: \: \: \: \: \dfrac{ \alpha }{2} \: \: \: \: \: \: \dfrac{ \alpha }{2}$

So , total moles at equilibrium be n ;

$\therefore \: n = (1 - \alpha ) + \dfrac{ \alpha }{2} + \dfrac{ \alpha }{2}$

$= > \: n = (1 - \alpha ) + \alpha$

$= > \: n = 1$

So, final answer is:

$\boxed{ \bf{\: n = 1}}$