Multiple Choice
When 5.60 g of solid sodium reacts with excess fluorine gas to form sodium fluoride, 13.8 kJ of heat is evolved at standard state conditions. What is the standard enthalpy of formation (ΔH⁰f in kJ/mol) of solid sodium fluoride?
8. Thermochemistry
Enthalpy of Formation
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Using the bond energies in Table 7.2 of your textbook, calculate the approximate enthalpy change for the following reaction: 2 H3C - CH3(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g). What is the enthalpy change?
A
-1400 kJ/mol
B
-700 kJ/mol
C
-2800 kJ/mol
D
-5600 kJ/mol
Verified step by step guidance1
Identify the bonds broken and formed in the reaction. For the reactants, you have C-C and C-H bonds in ethane (H3C-CH3) and O=O bonds in oxygen (O2). For the products, you have C=O bonds in carbon dioxide (CO2) and O-H bonds in water (H2O).
Use the bond energies from Table 7.2 to calculate the total energy required to break the bonds in the reactants. Multiply the bond energy of each type of bond by the number of such bonds in the reactants. For example, calculate the energy for breaking C-C, C-H, and O=O bonds.
Calculate the total energy released when the bonds in the products are formed. Again, use the bond energies from Table 7.2 and multiply by the number of each type of bond formed in the products, such as C=O and O-H bonds.
Determine the enthalpy change for the reaction by subtracting the total energy released (from forming bonds) from the total energy required (to break bonds). This is given by the formula: ΔH = Σ(bond energies of bonds broken) - Σ(bond energies of bonds formed).
Interpret the sign of the enthalpy change. If the result is negative, it indicates that the reaction is exothermic, meaning it releases energy to the surroundings.
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