1. When 100 cm of 1.00 mol dm sodium hydroxide, NaOH(aq), is added to 100 cm of 1.00 mol dm hydrochloric acid, HCl(aq), the temperature increases from 19.3 °C t

Réponse :

Explications :

■ NaOH + HCl --> H2O + NaCl + chaleur

■ en grammes :

  40       + 36,5 --> 18    + 58,5

■ en mole :

    1         +  1     -->   1     +    1

  concentration = 1 mole/dm³

  we use 100 cm³ --> donc 0,1 mole

■ enthalpy change of neutralization ( négative ! ) :

   we obtain 200 cm³ of water

  200 gram x 4,18 J/K.g x (26,1 - 19,3)°K / 0,1 mole

  = 56848 Joule / mole

  ≈ 57 kJ/mole  

■ remarque :

   the enthalpy change of neutralization is near - 58 kJ/mol .

   ( lachimie.org )