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Question: The standard enthalpies of the formation of \( C{O_2}(g) \) , \( {H_2}O(l) \) and glucose at \( {25^...

The standard enthalpies of the formation of CO2(g)C{O_2}(g) , H2O(l){H_2}O(l) and glucose at 25∘C{25^ \circ }C are −400KJ/mol,−300KJ/mol- 400KJ/mol, - 300KJ/mol , respectively. The respective. The standard enthalpy of combustion per gram of glucose at 25∘C{25^ \circ }C is :
(A) +2900KJ+ 2900KJ
(B) −2900KJ- 2900KJ
(C) −16.11KJ- 16.11KJ
(D) +16.11KJ+ 16.11KJ

Explanation

Solution

At first we will write what is given in the question. We will write the reaction of combustion of the glucose and produce the water and Carbon dioxide. Then we will write the net enthalpy equation. We will calculate the net enthalpy change of the reaction. Then we will calculate the standard enthalpy of the combustion by dividing the enthalpy change of molecular weight of glucose. We will find the correct option.

Complete step by step solution:
Step1. We have given the enthalpy formation of CO2(g)C{O_2}(g) is 400KJ/mol400KJ/mol
The enthalpy formation of H2O{H_2}O is −300KJ/mol- 300KJ/mol
The given temperature is 25∘C{25^ \circ }C . WE need to find the standard enthalpy of the combustion of the glucose.
Step2. When the reaction of the glucose is made with oxygen it produces the water and carbon dioxide. It also produces the energy. The reaction is given below
C6H12O6+6O2→6H2O+6CO2{C_6}{H_{12}}{O_6} + 6{O_2} \to 6{H_2}O + 6C{O_2}
Step3. The enthalpy change will be the difference of the enthalpy of the product and the reactants.
ΔH∘=∑ΔHf(saved)∘−∑ΔHf(reactan⁡t)∘ ∴ΔH∘=(6×(−400)+6×(−300))−(−1300+0) ⇒ΔH∘=−4200+1300=−2900KJ/mol  \Delta {H^ \circ } = \sum \Delta H_{f(saved)}^ \circ - \sum \Delta H_{f(reac\tan t)}^ \circ \\\ \therefore \Delta {H^ \circ } = (6 \times ( - 400) + 6 \times ( - 300)) - ( - 1300 + 0) \\\ \Rightarrow \Delta {H^ \circ } = - 4200 + 1300 = - 2900KJ/mol \\\
So we get eh enthalpy change here.
Step4. Now we will find the standard enthalpy by dividing the enthalpy change to the molecular weight of the glucose.
The molecular weight of the glucose is the 180gm180gm .
ΔH∘(KJ/mol)=−2900180=−16.11KJ/gm\Delta {H^ \circ }(KJ/mol) = \dfrac{{ - 2900}}{{180}} = - 16.11KJ/gm
Hence the standard enthalpy of combustion per gram will be −16.11Kj/mol- 16.11Kj/mol of glucose will be at 25∘C{25^ \circ }C .
Hence the correct answer will be option C.

Note:
The standard enthalpy of combustion is the change in enthalpy when one ole of the product burns during the standard condition and temperature. It is also known as heat of combustion. It is represented by the ΔH∘\Delta {H^ \circ }