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Total number of moles for the reaction `2Hl rarr H_(2) + I_(2)`. If `alpha` is degree of dissociation isA. 2B. `2 - alpha`C. 1D. `1 - alpha` |
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Answer» Correct Answer - A `{:("Initial",2HI,hArr,H_(2)+,I_(2)),(,,,0,0):}` At equilibrium `2 - alpha (alpha)/(2) (alpha)/(2)` Total mole `= 2 - alpha + (alpha)/(2) + (alpha)/(2) = 2`. |
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