The bond enthalpies for six selected chemical bonds are shown below.Bond Bond enthalpyΔH°298 (kJ/mol)H―H 436C―H 413C=O 532N―H 335O―H 499O=O 495For a reaction in which 2 moles of H2(g) reacts with 1 mole of O2(g) to form 2 moles of H2O(g), what is the heat of the reaction?A.1367 kJB.369 kJC.−629 kJD.−1996 kJ
Question
The bond enthalpies for six selected chemical bonds are shown below.Bond Bond enthalpyΔH°298 (kJ/mol)H―H 436C―H 413C=O 532N―H 335O―H 499O=O 495For a reaction in which 2 moles of H2(g) reacts with 1 mole of O2(g) to form 2 moles of H2O(g), what is the heat of the reaction?A.1367 kJB.369 kJC.−629 kJD.−1996 kJ
Solution
To calculate the heat of the reaction, we need to subtract the total bond enthalpy of the reactants from the total bond enthalpy of the products.
The reaction is: 2H2(g) + O2(g) -> 2H2O(g)
First, let's calculate the total bond enthalpy of the reactants:
- There are 2 moles of H2, and the bond enthalpy for H-H is 436 kJ/mol. So, the total bond enthalpy for H2 is 2 * 436 = 872 kJ.
- There is 1 mole of O2, and the bond enthalpy for O=O is 495 kJ/mol. So, the total bond enthalpy for O2 is 1 * 495 = 495 kJ.
- Therefore, the total bond enthalpy of the reactants is 872 + 495 = 1367 kJ.
Next, let's calculate the total bond enthalpy of the products:
- There are 2 moles of H2O, and each molecule of H2O has two O-H bonds. The bond enthalpy for O-H is 499 kJ/mol. So, the total bond enthalpy for H2O is 2 * 2 * 499 = 1996 kJ.
- Therefore, the total bond enthalpy of the products is 1996 kJ.
Finally, subtract the total bond enthalpy of the reactants from the total bond enthalpy of the products to find the heat of the reaction:
- ΔH = (total bond enthalpy of products) - (total bond enthalpy of reactants) = 1996 kJ - 1367 kJ = 629 kJ.
However, because the bonds are being broken in the reactants and formed in the products, the sign of the heat of the reaction is negative. Therefore, the heat of the reaction is -629 kJ.
So, the correct answer is C. -629 kJ.
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