What is the change in enthalpy (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if the bond enthalpies are: H-H = 436 kJ/mol, O=O = 498 kJ/mol, H-O = 463 kJ/mol?
Practice Questions
1 question
Q1
What is the change in enthalpy (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if the bond enthalpies are: H-H = 436 kJ/mol, O=O = 498 kJ/mol, H-O = 463 kJ/mol?
−484 kJ
−572 kJ
−572 kJ
−484 kJ
ΔH = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products) = [2(436) + 498] - [4(463)] = −572 kJ.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the change in enthalpy (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if the bond enthalpies are: H-H = 436 kJ/mol, O=O = 498 kJ/mol, H-O = 463 kJ/mol?
Step 1: Identify the bonds in the reactants and products. The reactants are 2H2 and O2, and the products are 2H2O.
Step 2: Count the number of each type of bond in the reactants. For 2H2, there are 2 H-H bonds. For O2, there is 1 O=O bond.
Step 3: Calculate the total bond enthalpy for the reactants. The bond enthalpy for H-H is 436 kJ/mol, and for O=O it is 498 kJ/mol. So, total for reactants = 2(436) + 498.
Step 4: Count the number of bonds in the products. For 2H2O, there are 4 H-O bonds (2 per H2O molecule).
Step 5: Calculate the total bond enthalpy for the products. The bond enthalpy for H-O is 463 kJ/mol. So, total for products = 4(463).
Step 6: Use the formula ΔH = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products) to find ΔH.
