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 $KCl$?

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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 $KCl$?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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:

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

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 $KCl$?

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