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Using the data in the table below calculate ΔE for the reaction Li(g) Cl(g)→Li (g) Cl−(g)

Using the data in the table below calculate ΔE for the reaction Li(g) Cl(g)→Li (g-example-1

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The change in energy for the reaction Li(g) + Cl(g) → Li+(g) + Cl-(g) is 171 kJ/mol.

There is the calculation of ΔE for the reaction Li(g) + Cl(g) → Li+(g) + Cl-(g) using the data in the table:

ΔE = ΔH = -IE - EA

Where:

ΔE is the change in energy

ΔH is the enthalpy change

IE is the ionization energy

EA is the electron affinity

For lithium:

IE = 520 kJ/mol

For chlorine:

EA = -349 kJ/mol

Now, we can plug these values into the equation:

ΔE = -520 kJ/mol - (-349 kJ/mol) = 171 kJ/mol

Therefore, the change in energy for the reaction Li(g) + Cl(g) → Li+(g) + Cl-(g) is 171 kJ/mol.

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