\begin{gathered}\text{At }444^\circ\mathrm C,\ \mathrm{HI}\text{ dissociates to the extent of 22.22% at equilibrium.}\[8pt]\text{What is the percentage dissociation of HI at 444°C, if the experiment

Question

\begin{gathered}\text{At }444^\circ\mathrm C,\ \mathrm{HI}\text{ dissociates to the extent of 22.22% at equilibrium.}\[8pt]\text{What is the percentage dissociation of HI at 444°C, if the experiment is}\\text{started with one mole of HI(g) and one mole of }\mathrm{I_2}\text{(g)?}\end{gathered}

TestHub · Accepted Answer

K c ​ = α 0 2 ​ / [ 4 ( 1 − α 0 ​ ) 2 ] = 1/49 , α 0 ​ = 2/9. Initially hydrogen is absent, so Q = 0 and HI dissociates forward. Starting with 1 mol HI and 1 mol iodine, at equilibrium: n HI ​ = 1 − α , n H 2 ​ ​ = α /2 , n I 2 ​ ​ = 1 + α /2. K c ​ = 4 ( 1 − α ) 2 α ( 2 + α ) ​ = 49 1 ​ . For fixed α > 0 this exceeds the pure-HI quotient, so α < 2/9. Dissociation is less than 22.22%: (a). ​

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