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`        when 3 mol of a and 1 mole of b are mixed in a 1 litre vessel the following reaction takes place a+b→2c    the equilibrium mixture contains 0.5 moles of c. what is the value of equilibrium constant for the reaction`
2 years ago

Arun
23330 Points
```							Note: we should always put the concentratin with their unit/(3-x)*(1-x)Put here the value of x and calculate K2/[A]*[B]=(2x)2please follow the answers and try to understand it properly, if still having some doubts then don’t hesitate in posting on the forum.in the given question, volume=1L, so initial concentration of A=3mol/1L=3MSimilarly for B=1MNow,A(g) + B(g)= 2C(g)3mol/L.....1mol/L..0.......(initially)3-x..........1-x …......2x...at equilibriumBut given that, mole of C=0.5 at equilibrium, Hence its equilibrium concentration=0.5/1L=0.5M=2x(from our equation)So x=0.25Mwriting for the equilibrium constantK=[C]  RegardsArun (askIITians forum expert)
```
2 years ago
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