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`        For the reactionSno2(s)+2H2(g) --->2H2o(g)+sn(l)Calculate ko at 900k ,where the equilibrium steam hydrogen mixture was 45% HE by volumeA)1.49B)1.22C).67D)none of these`
one year ago

Arun
23357 Points
```							Â In the reacn.Â Â  Â  Â  Â  Â  Â  Â  Â  SnO2(s) + 2H2(g) = 2H2O(g) + Sn(l)Since solid and liquid states compounds would not take part in Kc or Kp.Hence, from chemical eqn.,Â  Â  Â  Â  Â  Â  Â  Â  SnO2(s) + 2H2(g) = 2H2O(g) + Sn(l)Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  t=0 Â  Â  Â  Â V Â  Â  Â  Â  Â  Â  Â  Â  0Â Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  t=t Â  Â  Â V â€“ 2x Â  Â  Â  Â  Â  2xNow given => Â 2x = 0.45(V â€“ 2x)solving we get x = 0.45V/0.9 => V/2Hence Fnal volume becomes,Â Â at t=t Â  Â  Â  Â  Â  Â  Â  Kc = [(H2O)/(H2)]^2 = > [(2x)/(V-2x)]^2 = > 0.45^2 = > 20.25/100 = > 0.2025Â Kp = Kc*(RT)^nn = 3 â€“ 2 = 1Kp = 0.2025*0.0821*900 = >Â Â now you can solve this
```
one year ago
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