# Oxidation of glucose

By oxidation of 9 g of glucose it is released 150 kJ of heat. Calculate the standard enthalpy of derivation of glucose (in kJ/mol), if the given enthalpies of the reactants are ΔfH (H2O(l)) = -285,8 kJ/mol; ΔfH (CO2(g)) = -393,5 kJ/mol. Mr (glucose) = 180 (The answer is -1075.8)

$$\ce{6 CO2 + 6 H2O -> C6H12O6 + 6O2}$$

$$n(\ce{C6H12O6})=\frac{\pu{9g}}{\pu{180 g mol^-1}}=\pu{0.05 mol}~\ce{C6H12O6}$$

Now, $$\frac{\pu{0.05 mol}}{\pu{150 kJ}}=\frac{\pu{1 mol}}{x}\implies x=\pu{-3000 kJ/mol}$$

$$\Delta_\mathrm rH=\Delta_\mathrm fH(\ce{H2O(l)})+\Delta_\mathrm fH(\ce{CO2(g)})-\Delta_\mathrm fH(\ce{C6H12O6(s)})$$ $$= 6 \times (\pu{-285.8 kJ/mol}) + 6 \times (\pu{-393.5 kJ/mol}) - (\pu{-3000 kJ/mol}) = \pu{-1075.8 kJ/mol}$$

Did I take the right steps?

• "Did I take the right steps?" Alex, please note that "check my work" type questions are generally off-topic. Please highlight exactly which calculation step you are suspicious of, or feel would give the wrong answer. In any case, your answer matches the one given by the answer key. So, what do you feel is the problem here? Jun 22, 2018 at 17:09
• At first it seemed like I would get the wrong answer, as I was doing this problem in my notebook. However, when I wrote it here I got the right answer. Tbh, I had some difficulties figuring out what the reaction was, ie. what are the reactants and products of the reaction. I couldn't have figure it out myself until I found the reaction on the Internet.
– Alex
Jun 22, 2018 at 17:19
• Note that $n(\ce{C6H12O6})=\pu{0.05 mol}$ and not $n(\ce{C6H12O6})=\pu{0.05 mol}~\ce{C6H12O6}$.
– user7951
Jun 22, 2018 at 18:09

I had some difficulties figuring out what the reaction was

The question mentions "oxidation of glucose", that's a very important hint that tells you that the reaction is going to be:

$$\ce{C6H12O6 + O2 -> ?}$$

Now, note that oxidation of any carbohydrate/hydrocarbon/organic compound only yields (in general) carbon dioxide and water.

Thus, your complete reaction should be:

$$\ce{C6H12O6 + 6O2 -> 6 CO2 + 6 H2O}$$

Notice that each of the products is just the "oxide" of the elements of the original compound $\ce{(C, H)}$