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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..

Grade:11

1 Answers

Vikas TU
14149 Points
7 years ago
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           2x
Now given =>  2x = 0.4(V – 2x)
solving we get x = 0.4V/0.8 => V/2
Hence Fnal volume becomes, 
 
at t=t               Kc = [(H2O)/(H2)]^2 = > [(2x)/(V-2x)]^2 = > 0.4^2 = > 16/100 = > 0.16.
 
Kp = Kc*(RT)^n
n = 3 – 2 = 1
Kp = 0.16*0.0821*900 = > 11 (approx.)
(d) ption suits the best.

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