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Question: For the reaction, \(Sn{O_2}(s) + 2{H_2}(g) \to Sn(l) + 2{H_2}O(g)\) the equilibrium mixture of steam...

For the reaction, SnO2(s)+2H2(g)→Sn(l)+2H2O(g)Sn{O_2}(s) + 2{H_2}(g) \to Sn(l) + 2{H_2}O(g) the equilibrium mixture of steam and hydrogen contained 45%45\% and 24%24\% H2{H_2} at 900K900K and 1100K1100K respectively. Calculate KP{K_P} at both the temperatures. Generally should it be higher or lower temperatures for better reduction ofSnO2Sn{O_2} ?
A. ∴Kc2=1.03\therefore K{c_2} = 1.03
B. ∴Kc2=5.02\therefore K{c_2} = 5.02
C. ∴Kc2=15.01\therefore K{c_2} = 15.01
D. ∴Kc2=10.03\therefore K{c_2} = 10.03

Explanation

Solution

The mixture of steam and hydrogen is present in gaseous state only. The relation of KC{K_C} with the concentration is that the equilibrium constant is the ratio of concentration of product raised to its stoichiometric coefficient to the concentration of reactant raised to its stoichiometric coefficient.

Complete answer:
In the given reaction:
SnO2(s)+2H2(g)→Sn(l)+2H2O(g)Sn{O_2}(s) + 2{H_2}(g) \to Sn(l) + 2{H_2}O(g)
The solid Tin (IV) oxide is reduced with the help of hydrogen gas and produces molten Tin solution along with the steam of water. This reaction takes place at high temperature.
In the question, we have two cases:
Case I: The reaction that takes place at 900K.
SnO2(s)+2H2(g)→Sn(l)+2H2O(g)Sn{O_2}(s) + 2{H_2}(g) \to Sn(l) + 2{H_2}O(g)
(45)(45) (55)(55)
Thus, the first equilibrium constant is found to be:
Kc1=[H2O]2[H2]2=[55]2[45]2=1.494K{c_1} = \dfrac{{{{\left[ {{H_2}O} \right]}^2}}}{{{{\left[ {{H_2}} \right]}^2}}} = \dfrac{{{{\left[ {55} \right]}^2}}}{{{{\left[ {45} \right]}^2}}} = 1.494
Case II: The reaction that takes place at 1100K.
SnO2(s)+2H2(g)→Sn(l)+2H2O(g)Sn{O_2}(s) + 2{H_2}(g) \to Sn(l) + 2{H_2}O(g)
(24)(24) (76)(76)
Thus, the second equilibrium constant is found to be:
Kc2=[H2O]2[H2]2=[76]2[24]2=10.03K{c_2} = \dfrac{{{{\left[ {{H_2}O} \right]}^2}}}{{{{\left[ {{H_2}} \right]}^2}}} = \dfrac{{{{\left[ {76} \right]}^2}}}{{{{\left[ {24} \right]}^2}}} = 10.03
As we can observe that on increasing the temperature, the value of equilibrium constant increases rapidly. This proves that the reduction of SnO2Sn{O_2} takes place at a high temperature and the overall reaction is an endothermic reaction. On increasing the temperature of the reaction, the amount of product is more.

So, the correct answer is Option D.

Note:
The compound Tin (IV) oxide has strong covalent bonds in it which requires very high energy to break down. Thus, the reduction of this compound requires a very amount of energy as there is an attack of hydrogen atoms on the pi-bonds of the Sn=O bonds.