[Isomer]

If solid glucose is completely burned in the flame of a Bunsen burner, the enthalpy change is

A. greater than it is during cellular respiration because the production of water gas releases more energy than does the production of water liquid.

B. less than it is during cellular respiration because teh production of water gas releases less energy than does the production of water liquid.

I know the answer is B , but I'm thinking A.

I realize the glucose being burned is in an open environment, so the state of water vapor is a gas. In cellular resp the state of water is a liquid.

But I'm stumped at what this info is actually saying. Can somebody maybe explain it in laymans terms to me?

[Yggdrasil]

You can think of chemical equations kind of like regular mathematical equations and add them together.  So for example, if you add the metabolism of glucose in the cell and the vaporization of water:

C₆H₁₂O_6(s) + O_2(g) --> CO_2(g) + H₂O_(l)               (Delta)H₁
                     H₂O_(l) --> H₂O_(g)  (Delta)H₂
------------------------------------------
C₆H₁₂O_6(s) + O_2(g) --> CO_2(g) + H₂O_(g)  (Delta)H₃

(since H₂O_(l) appears on both sides of the resulting equation, you can cancel it out.  Note that the resulting equation is the equation of burining solid glucose in a bunsen burner)

When adding equations in this way, the changes in enthalpy are also additive, so:

(Delta)H₁ + (Delta)H₂ = (Delta)H₃

Since the vaporization of water is endothermic, (Delta)H₂ is positive and (Delta)H₁ > (Delta)H₃.

[note: the question is really an oversimplification of what happens in reality, while these answers are "correct", the real solution to the question would be a little more complicated]

[Isomer]

Are you referring to Hess's Law? 

[Yggdrasil]

Not exactly, but Hess's law is based on this principle.  Basically, the argument I provided below is valid because of the fact that enthalpy is a state function (i.e. it is path independent).