The enthalpy of dissociation for hydrogen
H2(g) → 2 H(g)
is ΔH = 436.0 kJ/mol. What is the difference between the bond enthalpy (ΔH) and bond energy (ΔE) for this process under standard conditions at 298.15 K?
To determine the difference between bond enthalpy (ΔH) and bond energy (ΔE) for the given process, we need to understand the definitions of these terms.
Bond enthalpy (ΔH) is the change in enthalpy when a particular bond in a compound is broken or formed, while bond energy (ΔE) is the amount of energy required to break a specific bond in a molecule.
Typically, bond enthalpy is determined experimentally by measuring the heat released or absorbed during a chemical reaction. On the other hand, bond energy is a theoretical value calculated using quantum mechanical calculations.
In the given reaction H2(g) → 2 H(g), the enthalpy change (ΔH) is given as 436.0 kJ/mol. This represents the amount of heat released or absorbed when one mole of H2(g) is converted into two moles of H(g).
To find the difference between bond enthalpy (ΔH) and bond energy (ΔE), we need to consider the stoichiometric coefficients of the reactants and products.
In this case, one mole of H2(g) produces two moles of H(g), indicating that the bond enthalpy of H2(g) is split into two H(g) bonds.
Therefore, the bond enthalpy (ΔH) of the H-H bond in H2(g) can be divided by 2 to obtain the bond enthalpy of one H-H bond.
ΔH = 436.0 kJ/mol
ΔH(H-H) = ΔH/2 = 436.0 kJ/mol / 2 = 218.0 kJ/mol
Now, bond energy (ΔE) can be calculated by multiplying the bond enthalpy by the Avogadro's number (6.022 x 10^23 mol^-1) to convert it into energy per molecule.
ΔE = ΔH(H-H) x (1 mol / 6.022 x 10^23 mol^-1) = 218.0 kJ/mol / 6.022 x 10^23 mol^-1 ≈ 3.623 x 10^-19 J/molecule
Therefore, the difference between bond enthalpy (ΔH) and bond energy (ΔE) for the dissociation of hydrogen gas is approximately 3.623 x 10^-19 J/molecule under standard conditions at 298.15 K.
The bond enthalpy (ΔH) is the amount of energy required to break a specific bond in a molecule, while the bond energy (ΔE) is the average energy of a bond in a sample of molecules.
In the given reaction, the dissociation of hydrogen molecule (H2) into two hydrogen atoms (2H) is exothermic, meaning energy is released during the process. The enthalpy change (ΔH) is -436.0 kJ/mol, indicating that 436.0 kJ of energy is released per mole of reactant.
Under standard conditions at 298.15 K, the bond energy (ΔE) can be calculated by adding the bond enthalpy (ΔH) to the product of the gas constant (R) and the temperature (T):
ΔE = ΔH + RT
Where:
ΔE = Bond energy
ΔH = Bond enthalpy
R = Gas constant (8.314 J/mol·K)
T = Temperature (298.15 K)
Substituting the values:
ΔE = -436.0 kJ/mol + (8.314 J/mol·K)(298.15 K)
= -436.0 kJ/mol + 2480.36 J/mol
= -436.0 kJ/mol + 2.48036 kJ/mol
= -433.51964 kJ/mol
Therefore, the difference between bond enthalpy (ΔH) and bond energy (ΔE) for the process of hydrogen dissociation at standard conditions is approximately -433.52 kJ/mol.