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For the reaction SnO2(s) + 2H2(g) = 2H2O(g) + Sn(l) at 900 K the equillibrium steam hydrogen mixture was found to be 40% H2 by volume. The Kp is(a) 1.15 (b) 2.25 (c) 7.75 (d) 10

For the reaction SnO2(s) + 2H2(g) = 2H2O(g) + Sn(l) at 900 K the equillibrium steam hydrogen mixture was found to be 40% H2 by volume. The Kp is(a) 1.15 (b) 2.25 (c) 7.75 (d) 10

Grade:11

2 Answers

Vikas TU
14149 Points
7 years ago
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.
 
Md Shayan
13 Points
2 years ago
SnO2 + 2H2 gives 2H2O + Sb 
Kp= p[h2o]²/ p[H2]²
     = 60%/40% = 3/2 
( as 40% of H2 is already given ) 
Kp = 3²/2² = 9/4 = 2.2something.✅
The correct option which suits better is (b) 2.25 

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