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`        A+B=3C.at initial 2moles of A and 2 moles of B present at 1Liter contains .if mole fraction of C is 0.5 at equilibrium. Calculate equilibrium constant.`
2 years ago

```							Put here the value of x and calculate K/(2-x)*(2-x)^3Hello student!!please 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=2mol/1L=2MSimilarly for B=2MNow,A(g) + B(g)= 3C(g)2mol/L.....2mol/L..0.......(initially)2-x..........2-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]2/[A]*[B]=(2x) RegardsArun (askIITians forum expert)
```
2 years ago
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