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For the reaction SnO2(s) +2H2(g)= 2H2O(g)+ Sn(l) at 900K, the equillibrium steam hydrogen mixture was found to be 40% H2 by volume. The Kp is.. For the reaction SnO2(s) +2H2(g)= 2H2O(g)+ Sn(l) at 900K, the equillibrium steam hydrogen mixture was found to be 40% H2 by volume. The Kp is..
Re – Posting Again,As U have asked twice...In the reacn. 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., SnO2(s) + 2H2(g) = 2H2O(g) + Sn(l) t=0 V 0 t=t V – 2x 2xNow given => 2x = 0.4(V – 2x)solving we get x = 0.4V/0.8 => V/2Hence Fnal volume becomes, at t=t Kc = [(H2O)/(H2)]^2 = > [(2x)/(V-2x)]^2 = > 0.4^2 = > 16/100 = > 0.16. Kp = Kc*(RT)^nn = 3 – 2 = 1Kp = 0.16*0.0821*900 = > 11 (approx.)(d) ption suits the best.
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