energetics 問題

Question

我想問:

Given: H+ + OH- --->H2O enthalpy change = - 57kJmol^-1

calculate the enthalpy changes which would occur when the following solutions react as completely as possible:

(a)50cm3 of 0.01m Ca(OH)2 solution and 25cm3 of 0.01M HCl;
(b)50cm3 of 0.01m Ca(OH)2 solution and 100cm3 of 0.01M HCl;
(c)50cm3 of 0.01m Ca(OH)2 solution and 200cm3 of 0.01M HCl;

THX!~

Answer 1

Basic knowledge :

H+(aq) + OH-(aq) → H2O(l)
mole ratio H+(aq) : OH-(aq) : H2O(l) = 1 : 1 : 1

Enthalpy change of neutralization is defined as the enthalpy change in the standard state when 1 mole of water is produced from the neutralization of an acid and an alkali.

Each mole of Ca(OH)2 contains 2 moles of OH-(aq) ions, and each mole of HCl give off 1 mole of H+(aq) ions in water.

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1.
No. of moles OH-(aq) ion = 0.01 x (50/1000) x 2 = 0.001 mol
No. of moles H+(aq) ion = 0.01 x (25/1000) = 0.00025 mol
OH-(aq) ion is in excess.
No. of moles of H2O formed = 0.00025 mol
Enthalpy change = -57 x 0.00025 = -0.01425 kJ = -14.25 J

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2.
No. of moles OH-(aq) ion = 0.01 x (50/1000) x 2 = 0.001 mol
No. of moles H+(aq) ion = 0.01 x (100/1000) = 0.001 mol
OH-(aq) and H+(aq) ions are completely neutralized.
No. of moles of H2O formed = 0.001 mol
Enthalpy change = -57 x 0.001 = -0.057 kJ = -57 J

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3.
No. of moles OH-(aq) ion = 0.01 x (50/1000) x 2 = 0.001 mol
No. of moles H+(aq) ion = 0.01 x (200/1000) = 0.002 mol
H+(aq) ion is in excess.
No. of moles of H2O formed = 0.001 mol
Enthalpy change = -57 x 0.001 = -0.057 kJ = -57 J