Question

When `1.0` `g` of potassium chloride, `KCl`, is dissolved in `25` `mL` (=`25` `g`) of water, it causes the temperature to drop from `24.33^o` `C` to `22.12^o` `C`. What is the molar enthalpy of dissolution for `KCl`?

Answer 1

The energy absorbed from the water by this endothermic reaction is approximately `Delta H=17.3` `kJmol^-1`.

Explanation:

The reaction in question is:

`KCl_"(s)" to K^"+"(aq) + Cl^"-" (aq)`

The reaction is endothermic, so the value of `Delta H` we calculate at the end will be positive. First, though, we need to calculate the amount of energy absorbed from the water, which will be negative:

`DeltaH=mCDeltaT=mC(T_2-T_1)`
`=25xx4.186xx(22.12-24.33)=-232.05` `J`

This is the amount of energy absorbed from the water by the process of dissolving the `KCl`. This is the same as the energy gained by the 'system' of dissolving `KCl`. That means energy gained of `232.05` `J` for the `1.0` `g` sample. What we want to know, though, is the total energy absorbed when 1 mole of `KCl` dissolves.

One gram of KCl is: `n=m/M=1/74.55=0.0134` `mol`

That means the total energy gain if we had dissolved an entire mole, `74.55` `g`, of `KCl`, is 74.55 times as great, which gives:

`Delta H=74.55` `mol^-1xx232=17295.6` `Jmol^-1 ~~ 17.3` `kJmol^-1`