Why reaction between alkali metals and water exothermic?

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SUMMARY

The reaction between alkali metals and water is exothermic, releasing significant heat due to the strong hydration of cations formed during the reaction. The balanced chemical equation is 2M(s) + 2H2O(l) => 2M+(aq) + 2OH−(aq) + H2(g). The net electrochemical potential (Enet) for sodium, calculated as 1.88 volts, indicates a highly favorable reaction. The Gibbs free energy change (∆G) is -363 kJ, confirming the exothermic nature of the process, which aligns with Hess's Law and heat of formation data.

PREREQUISITES
  • Understanding of exothermic and endothermic reactions
  • Familiarity with electrochemical potentials and Gibbs free energy
  • Knowledge of ionization and hydration processes
  • Basic grasp of Hess's Law and heat of formation calculations
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  • Study the principles of electrochemistry, focusing on standard electrode potentials
  • Explore the concept of hydration enthalpy and its impact on reaction thermodynamics
  • Investigate Hess's Law applications in thermochemical calculations
  • Learn about the properties and reactions of alkali metals with water
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Chemistry students, educators, and researchers interested in thermodynamics and reaction kinetics, particularly those focusing on alkali metal reactions and electrochemical processes.

terryds
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Why the reaction between alkali metals and water exothermic??
I know that it produces heat to surroundings
But, what actually happens?? Why it gives heat to surroundings??

http://antoine.frostburg.edu/chem/senese/101/thermo/faq/exothermic-endothermic-examples.shtml says that when forming ion pairs (like ionization reaction), it is endothermic (since it needs some heat to get the electron out)

But, the reaction between alkali metals and water is

2M(s) + 2H2O(l) => 2M+(aq) + 2OH−(aq) + H2(g)

Which forms ions pair..
But, it's not endothermic, but exothermic.. Why?
 
Chemistry news on Phys.org
Cation hydration is strongly exothermic.
 
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2Nao(s) <=> 2Na+(aq) + 2e- Eo = -2.71 volts
H2O(l) + 2e- <=> 2OH-(aq) + H2(g) Eo = -0.83 volts
2Nao + 2H2O(l) <=> 2Na+(aq) 2OH- + H2(g) Enet = (Eredn - Eoxdn = [(-0.83v) - (-2.71v) = 1.88v (net)

∆ G = - nFEo = -[(2e-)(96,500C)(1.88v) = -362,840 joules = - 363 Kj

Compares well with Heat of Formation Tables using Hess's Law
ΔHrxn = [∑ΔHf(P) - ΣΔHf(R)] = [2mole(-240.1Kj/mole) + 2(-230.0Kj/mole)] - [-(2(285.8 Kj/mol)] = - 368.6 Kj
4
A common 1" diameter Parraffin wax candle will gives off about 4x more heat than in this lab study;i.e., ~ 1.2 x 10 4 Kj/mole wax.

 

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