Asked by bob
In general, which reaction(forward, reverse, or neither) is favored if the value of K at a specified temperature is
a. equal to 1?
b. very small
c. very large
a. equal to 1?
b. very small
c. very large
Answers
Answered by
bob
i was thinking of b
Answered by
bobpursley
Think again. What does K mean?
Answered by
bob
kelvin so c
Answered by
DrBob222
That answer is crazy. Keq is NOT Kelvin. And it appears to me that your question actually is three questions. They want to know three separate answers for a,b, and c.
Answered by
tommy
so how do you answer it
Answered by
tommy
i have this same question and I am so losed
Answered by
DrBob222
For a reaction, such as
A + B ==> C + D
Keq = (C)*(D)/(A)*(B)
So for K to be small, the numerator must be small and/or the denominator must large (that's the only way to get a small number out of a fraction), so C and D must be small and/or A and B large; therefore, the reaction does not proceed to a large extent. That is, A and B simply do not react to a great extent. For Keq to be large, then the numerator must be large and/or the denominator must be small which means the products over shadow the reactants and the reaction proceeds. That is, the products are favored over the reactants. For Keq = 1, products must be equal to reactants and the reaction is about 50-50.
A + B ==> C + D
Keq = (C)*(D)/(A)*(B)
So for K to be small, the numerator must be small and/or the denominator must large (that's the only way to get a small number out of a fraction), so C and D must be small and/or A and B large; therefore, the reaction does not proceed to a large extent. That is, A and B simply do not react to a great extent. For Keq to be large, then the numerator must be large and/or the denominator must be small which means the products over shadow the reactants and the reaction proceeds. That is, the products are favored over the reactants. For Keq = 1, products must be equal to reactants and the reaction is about 50-50.