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3.2 moles of HI(g) were heated in a sealed bulb at 444°C till the equilibrium was reached. It's 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 equilibrium constant for the reaction 2HI(g) H2(g)+I2(g). Considering the volume of the container 1L.
plz answer me soon and explain

Concentration of HI is 3.2 moles/1litre
=3.2M
degree of dissociation= 20%
degree of dissociation = no of moles of reactants dissociated/no. of moles of reactants present initially
20/100=x/3.2
x=0.64 moles
according to the reaction
2HI(g)=H2(g)+I2(g)
two M of HI gives one M of H2 and I2 each so .64 M of HI would give .32 M of H2 and I2 each
hence equilibrium concentration of HI is 2.56M
Kc=[0.32][0.32]/[2.56]2
=0.0156
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hey Shubhanshi how did you get 2.56??
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Given that, Initial conc. Of HI is 3.2 (since conc =n/v and here in the ques it's been givn that v=1. So, conc. = no. of moles). 20% of HI is dissociated & reaches equil. Therefore, At equil. [HI]= 2.56. {You'll get the no. of moles from here since I mentioned above that no. of moles = conc.}{3.2-0.64}. [HI]=[I2]=0.32 {0.64÷2}. Get the value of Kc. :)
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Answer is in picture

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3.2-x=? : 3.2- 0.64= 2.56
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2.56 is obtained by 3.2 - 0.64 which is number of moles left at equilibrium.
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This is right

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kc is 0.0156
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Hye guys Have a look on my solution Concentration of HI is 3.2 moles/1litre =3.2M degree of dissociation= 20% degree of dissociation = no of moles of reactants dissociated/no. of moles of reactants present initially 20/100=x/3.2 x=0.64 moles according to the reaction 2HI(g)=H2(g)+I2(g) two M of HI gives one M of H2 and I2 each so .64 M of HI would give .32 M of H2 and I2 each hence equilibrium concentration of HI is 2.56M Kc=[0.32][0.32]/[2.56]2 =0.0156
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This is the way

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Here's it

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How you calculate the value of no. Of moles of reactant present initially
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fygvuj
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hgjyj
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Here

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Please find the attached solution
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In this question how we get answer (3)? What these options signify plz tell anyone

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FIRST LET US UNDERSTAND THE CHEMICAL REACTION BY WRITING EQUATIONS

                          2HI                        ↔              H2       +            I2 
at t=0              3.2 moles                                0                        0
at t=eq.            3.2-2x                                     x                        x
  Now degree of dissociation is 20%
        Therefore 2x=(20/100)*3.2
                             =0.64
at eq
HI = 3.2-0.64=2.56 moles
H2= 0.32 moles
I2=0.32 moles
 
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Concentration of HI is 3.2 moles/1litre?
=3.2M?
degree of dissociation= 20%?
degree of dissociation = no of moles of reactants dissociated/no. of moles of reactants present initially?
20/100=x/3.2?
x=0.64 moles?
according to the reaction?
2HI(g)=H2(g)+I2(g)?
two M of HI gives one M of H2 and I2 each so .64 M of HI would give .32 M of H2 and I2 each?
hence equilibrium concentration of HI is 2.56M?
Kc=[0.32][0.32]/[2.56]2?
=0.0156
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If you are taking 0.32 moles of H2 and I2 both, in the equation then why is 2.56 mole taken for HI and why not 0.64
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Hope it helps??

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Tall me 7 lesson sir
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