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#?

1 Answer
Apr 10, 2016

Answer:

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#