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If alpha is the fraction of HI dissociat...

If `alpha` is the fraction of HI dissociated at equilibrium in the reaction, `2HI(g)hArrH_2(g)+I_2(g)` starting with the 2 moles of HI. Then the total number of moles of reactants and products at equilibrium are

A

`1.0`

B

`1+alpha`

C

`2.0`

D

`2+2alpha`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(,2HI,hArr,H_(2),+,I_(2)),("Inital",2,,0,,0),("At eqm.",2-alpha,,(alpha)/(2),,(alpha)/(2)):}`
`2-alpha+(alpha)/(2)+(alpha)/(2)=2`
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Knowledge Check

  • If alpha is the fraction of HI dissociated at equlibrium in the reaction :2H hArr H_(2)+I_(2) then starting with 2 mol of HI, the total number of moles of reactants and products at equlibrium are

    A
    1
    B
    2
    C
    `1+alpha`
    D
    `2+2alpha`

  • For the reaction , H_(2) + I_(2) hArr 2 HI , K= 47*6. If the intial number of moles of each reactant and product is 1 mole , then at equilibrium

    A
    `[I_(2)] = [H_(2)],[I_(2)] gt [HI]`
    B
    `[I_(2)]lt [H_(2)], [I_(2)] = [HI]`
    C
    `[I_(2)] = [H_(2)] , [I_(2)] lt [HI]`
    D
    ` [I_(2)] gt [H_(2)] , [I_(2)] = [HI]`

  • For the reaction, H_(2) + I_(2)hArr 2HI, K = 47.6 . If the initial number of moles of each reactant and product is 1 mole then at equilibrium

    A
    `[I_(2)] = [H_(2)], [I_(2)] gt [HI]`
    B
    `[I_(2)] = [H_(2)], [I_(2)] lt [HI]`
    C
    `[I_(2)] lt [H_(2)], [I_(2)] = [HI]`
    D
    `[I_(2)] gt [H_(2)], [I_(2)] = [HI]`
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