Chemistry - Thermochemistry Question with Solution | TestHub
The thermochemical equation for the dissociation of hydrogen gas into atoms may be written as: H2 → 2H; ΔH = 432.0 kJ. What is the ratio of the energy yield on combustion of hydrogen atoms to steam to the yield on combustion of an equal mass of hydrogen molecules to steam? Heat of formation of steam is -240.0 kJ/mol.
Options:
Answer:
Solution:
Consider 2 g of hydrogen.
When hydrogen reacts as molecules:
1 mol H2 + 1/2O2 tarrow H2O
The energy released is 240 kJ.
When hydrogen reacts as atoms:
2 mol H + 1/2O2 tarrow H2O
This reaction can be broken down into two steps:
1. The formation of hydrogen molecules from hydrogen atoms:
2H tarrow H2
The enthalpy change for this step is Δ H = -432 kJ.
2. The reaction of hydrogen molecules with oxygen:
H2 + 1/2O2 tarrow H2O
The enthalpy change for this step is -240 kJ.
The total energy released when hydrogen reacts as atoms is the sum of the absolute values of the enthalpy changes for these two steps:
Energy = |-432 kJ| + |-240 kJ| = 432 kJ + 240 kJ = 672 kJ.
The ratio of energy released when hydrogen reacts as atoms versus as molecules is:
Ratio = 672 kJ/240 kJ = 2.80.
This corresponds to option (a).
