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25 ml of H2 and 18 ml of I2 vapours were heated in a sealed tube at 456^. C,when at equilibrium 30.8 ml of HI was formed.Calculate the egree of dissociation of pure HI at 456^.C.

(ans is 24.5%)

r.d007

ans is 24.5%

akhil_o

$H_2+I_2\rightarrow2HI\\Amount\,of\,HI\,at\,equilibrium\,=30.4mL\\Since\,molar\,ratio\,is\,1:1:2\\The\,amount\,of\,H_2\,and\,I_2\,reacted\,is\,\\30.4/2=15.4mL$

$So\,at\,equilibrium\,we\,have\,\\H_2=25-15.4=9.6mL\\Cl_2=18-15.4=2.6mL\\So \,K_c= \frac {[HI]^2}{[H_2][I_2]}\\K_c=\frac{30.8^2}{9.6*2.6}$

$For\,reverse\,reaction\,K"=1/K_c\\Putting\,values\,K"=0.104\\Now \,Dissociation\,of\,HI\,\\K"=\frac{[\alpha]^2}{(2(1-\alpha))^2}\\0.104=(\alpha^2)/(4(1-\alpha ^2))$

$Solving \,for\,alpha\,we\,get\,\\ \alpha=0.244\\Percentage\,dissociation=24.4$

allamraju

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I too tried the same way but got a diff ans.May be I made a calculation mistake somewhere.Anyway,the ans given by akhil is perfect.