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The equilibrium constant Kp for the reaction 2H2O(g)--> 2H2(g)+O2(g) is 2x10^-42 at 25 degrees C. (a)...

The equilibrium constant Kp for the reaction 2H2O(g)--> 2H2(g)+O2(g) is 2x10^-42 at 25 degrees C. (a) what is Kc for the reaction at the same temperature? (b) the very small value of Kp (and Kc) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

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Answer #1

WE know that the relation between Kp and Kc is

Kp = Kc (RT)^ng

ng = Number of moles of gaseous products - Number of moles of gaseous reactants

ng = (2+1)-2 = 1

Kp = Kc (RT)^1

Kc = Kp / (RT)

R = 0.0821

T = 25 C = 298 K

Kc = 2 X 10^-42 / 0.0821 X 298 = 0.0817 X 10^-42 = 8.17 X 10^-44

b) The reaction is thermodynamically favourable however kinetically it is not favorable. It has a very low rate of reaction

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