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)


$$\ce{6 CO2 + 6 H2O -&gt; C6H12O6 + 6O2}$$

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

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?
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:

$$\ce{C6H12O6 + O2 -&gt; ?}$$

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

Thus, your complete reaction should be:

$$\ce{C6H12O6 + 6O2 -&gt; 6 CO2 + 6 H2O}$$

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