Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl

1.	Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl in an excess of water. Does this process represent an ionization reaction? Explain.ΔH°f[K+(aq)] = -251.2 kJ mol-1ΔH°f[Cl-(aq)] = -167.08 kJ mol-1ΔH°f[KCl] = -437.6 kJ mol-1
Answer» ΔH = ΔH°_fK⁺(aq) + ΔH°_fCl^-(aq) - ΔH°_fKCl(s) = -251.2 - 167.08 - (-437.6) = -418.28 + 437.6 = 19.32 kJ The reaction represents a solution process and not an ionisation. The KCl solid is ionic before being dissolved as well as after.

Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl in an excess of water. Does this process represent an ionization reaction? Explain.ΔH°f[K+(aq)] = -251.2 kJ mol-1ΔH°f[Cl-(aq)] = -167.08 kJ mol-1ΔH°f[KCl] = -437.6 kJ mol-1

Answer»

ΔH = ΔH°_fK⁺(aq) + ΔH°_fCl^-(aq) - ΔH°_fKCl(s)

= -251.2 - 167.08 - (-437.6)

= -418.28 + 437.6

= 19.32 kJ

The reaction represents a solution process and not an ionisation. The KCl solid is ionic before being dissolved as well as after.

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Calculate ΔH for the process at 25°C of dissolving 1.00 mol of KCl in an excess of water. Does this process represent an ionization reaction? Explain.ΔH°f[K+(aq)] = -251.2 kJ mol-1ΔH°f[Cl-(aq)] = -167.08 kJ mol-1ΔH°f[KCl] = -437.6 kJ mol-1

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