Question

4.954g (6.645x10^-2 moles) of KCl dissolved in 40g of H2O. Initial temp of salt/water 24.4 degC. Final temp of salt/water 18.7 degC.

how do i find Enthalpy of formation of the anion and cation?

Answer 1

use:

Delta H= -m*C* (T2-T1)

Here c is specific heat capacity of water,m is mass of water and T is temperature

putting values,

Delta H= -m*C* (T2-T1)

     = - 40 * (4.184)*(18.7 - 24.4)

    = 954 J

Enthalphy of formation of cation or anion is defined as change in heat when 1 mol of substance is dissolved. So we need to calculate heat change when 1 mol of KCl is dissolved.

when 6.645x10^-2 moles are dissolved delta H = 954 J

When 1 mol will be dissolved, delta H = 954 / (6.645x10^-2) = 14357 J = 14.36 KJ

Answer: 14.36 KJ