Some bond enthalpies are given - AQA - A-Level Chemistry - Question 10 - 2020 - Paper 3

Next, we calculate the total bond enthalpy for the products. For 1 molecule of CO₂, there are 2 C=O bonds, and for 2 molecules of H₂O, there are 4 O–H bonds:

$ext{Total bond enthalpy for products} = 2 \times 743 \,\text{kJ} + 4 \times 463 \,\text{kJ}$

Calculating this gives:

$= 1486 + 1852 = 3338 \,\text{kJ}$

The enthalpy change (\Delta H) for the reaction can be found using the formula:

$\Delta H = \text{Total bond enthalpy for products} - \text{Total bond enthalpy for reactants}$

Substituting in the values gives:

$\Delta H = 3338 - 2640 = 698 \,\text{kJ}$

Since the value is positive, the reaction is endothermic. Thus, the answer is option D: -698 kJ.