# Chemistry - What does K° signify in this reaction?

## Solution 1:

K° here means elemental potassium (zero oxidation state). The author wanted to avoid the confusion of K with the equilibrium constant. There are more quantities than we have symbols, so letters and symbols get re-used. It is called a pigeon hole effect.

Now, that the OP has posted the reference, Solomons, Organic Chemistry, Ed 11, Problem 11.14 Solution manual, it is clear that the circle is a typo for zero. In the 11 and the 12th edition, the symbol stands corrected to proper numerical "zero" in the main textbook (pg 510). It is not unusual to find typos in solution manuals and textbooks. We are all humans-after all.

## Solution 2:

It is possible that author's intent was to denote that the reaction with potassium metal is carried out under ambient conditions. One could use a superscript circle to refer to the standard state of a pure substance, and even though only the pressure is standardized, the notation of standard state is often utilized for denoting a temperature of 25 °C.

This is non-conventional use of a standard state notation and should not be encouraged.

Why it's probably not a notation for a zerovalent metal? First, a zerovalency is denoted with, well, zero, not a circle. Second, it appears that the image for the reaction scheme was produced using a $$\mathrm\LaTeX$$ compiler, and usually those who are using such typesetting system know the difference between zero “0” and a circle “○”. One does not occasionally type K^\circ $$(\ce{K^\circ})$$ in place of K^0 $$(\ce{K^0}).$$

Why it's probably not an equilibrium constant? Balance aside, it could be a standard equilibrium constant $$K^\circ,$$ but since the symbol is upright $$(\ce{K^\circ})$$ and not slanted/italicized, it eliminates this possibility. There is no need for extra symbols to distinguish between the symbols for an element and for a variable. Same goes, e.g. for phosphorus $$\ce{P}$$ and power $$P,$$ or nitrogen $$\ce{N}$$ and number of particles $$N.$$