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For the reaction Sno2(s)+2H2(g) --->2H2o(g)+sn(l) Calculate ko at 900k ,where the equilibrium steam hydrogen mixture was 45% HE by volume A)1.49 B)1.22 C).67 D)none of these

For the reaction
Sno2(s)+2H2(g) --->2H2o(g)+sn(l)
Calculate ko at 900k ,where the equilibrium steam hydrogen mixture was 45% HE by volume
A)1.49
B)1.22
C).67
D)none of these

Grade:12th pass

1 Answers

Arun
25750 Points
5 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 = 0.45(V – 2x)
solving we get x = 0.45V/0.9 => V/2
Hence Fnal volume becomes, 
 
at t=t               Kc = [(H2O)/(H2)]^2 = > [(2x)/(V-2x)]^2 = > 0.45^2 = > 20.25/100 = > 0.2025
 
Kp = Kc*(RT)^n
n = 3 – 2 = 1
Kp = 0.2025*0.0821*900 = > 
 
now you can solve this

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