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For the reaction :P==Q+R.Initially 2 moles of P was taken .Up to equibrium 0.5 moles of P was dissociated .What would be the degree of dissociation

Asked by m.nilu 23rd September 2018, 11:00 AM
Answered by Expert
Answer:
Given:
 
Initial moles of P = 2 mole
 
Moles dissociated upto equilibrium = 0.5 moles
 
The equation is;
 
space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight P space space leftwards harpoon over rightwards harpoon space space space space space straight Q space space space space plus space space space space straight R space

space space space space space space space space space space Initial space space space space space space space space space space 2 space space space space space space space space space space space 0 space space space space space space space space space space space space 0

At space equilibrium space space space open parentheses 2 minus 2 straight alpha close parentheses space space space space space space 2 straight alpha space space space space space space space space space space 2 straight alpha

0.5 space moles space of space straight P space as space dissociated

therefore space 2 straight alpha space equals space 0.5

space space space space space space straight alpha space space equals space 0.25
 
The degree of dissociation is 0.25
Answered by Expert 24th September 2018, 2:51 PM
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