What is the significance of a very large Kp value on the equilibrium concentration of H2 in the reaction : 2H2 (g) + O2 (g) <-> 2H2O (g) ?

Question

A. There is very little H2 at equilibrium.
B. There is a significant amount of H2 at equilibrium.
C. The amount of H2 does not depend on the value of Kp
D. The amount of H2 is always equal to the amount of O2 at equilibrium.

I am pretty sure that the answer has to be either A or B as C and D don't really make sense. Can someone please help me out?

Answer 1

You're right. Neither C nor D are correct. So now you're down to guessing for A or B. But why guess? Write the Kp expression.
Kp = very large = p^2(H2O)/p^2(H2)*p(O2)
To get a very large number in ANY fraction you must have either a large numerator OR a very small denominator (or both). Now, shouldn't that tell you the there must be a bunch of products (the numerator) and a very small amount of reactants (the denominator)? Now you can answer the question. I just gave you the answer.

Answer 2

So the answer is A

Answer 3

yes. There will relatively large amount of H2O and relatively small amounts of H2 and O2 at equilibrium when Kp is large. The reverse is true when you have a small value for Kp. Have you studied Ksp yet? For solubility product, the numbers tell you how soluble a substance is. For example, AgCl has a Ksp of about 1E-10. AgCl(solid) ==> Ag^+(aq) + Cl^-(aq)
So Ksp expression is 1E-10 = (Ag^+)(Cl^-). With a number small like 1E-10 you know that AgCl is sparingly soluble and it takes little to form a saturated solution of AgCl.

Answer 4

Thank you DrBob222