Find the appropriate equilibrium constant (Kp) for the reaction below under the

Question

Find the appropriate equilibrium constant (Kp) for the reaction below under the following conditions.

1/2 SNO2(s) + H2(g) <--> H2O(g) + 1/2 Sn(s)

a. At 650*C, The equilibrium steam-H2 mixture was 38% H2 by Volume.
b. At 800*C, The equilibrium steam-H2 mixture was 22% H2 by Volume.
c. Would you run this reaction at a high or low temperature for the most efficient reduction of Tin?

Please show step by step! I have an icetable for the first one kinda..but I don't know how to get anything out of the %. I am so confused. I know you don't include solids for a Keq though..please help..

Explanation / Answer

ANSWER : THE GIVEN REACTION IS AS FOLLOWS

1/2 SNO2(s) + H2(g) <--> H2O(g) + 1/2 Sn(s)

1) Let us consider hydrogen volume = 0.38 V , (where V is the total volume)

the rest is h20 = 0.62 V

now we use the formula = kp = [pH2O ] / [pH2] = p [xH2O ] /  p [xH2]

Kp = 0.62 /0.38 = 1.63

So the Kp for this reaction is given by ratio of mole fractions .

2) Now for the second part we do it just like above

Let us consider hydrogen volume = 0.22 V , (where V is the total volume)

the rest is h20 = 0.78 V

now we use the formula = kp = [pH2O ] / [pH2] = p [xH2O ] /  p [xH2]

Kp = 0.22 /0.78 = 3.5

3) i will run this reaction at a high temperature for the most efficient reduction of Tin as more the temperature more will be the energy of reactants and thus they combine with each other to release energy to get stability .

The Kp increases for high temperature , reaction is more efficient for high temperature .

So the Kp for this reaction is given by ratio of mole fractions .

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