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?

Apr 10, 2016

The energy absorbed from the water by this endothermic reaction is approximately $\Delta H = 17.3$ $k J m o {l}^{-} 1$.

Explanation:

The reaction in question is:

$K C {l}_{\text{(s)" to K^"+"(aq) + Cl^"-}} \left(a q\right)$

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:

$\Delta H = m C \Delta T = m C \left({T}_{2} - {T}_{1}\right)$
$= 25 \times 4.186 \times \left(22.12 - 24.33\right) = - 232.05$ $J$

This is the amount of energy absorbed from the water by the process of dissolving the $K C l$. This is the same as the energy gained by the 'system' of dissolving $K C l$. 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 $K C l$ dissolves.

One gram of KCl is: $n = \frac{m}{M} = \frac{1}{74.55} = 0.0134$ $m o l$

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

$\Delta H = 74.55$ $m o {l}^{-} 1 \times 232 = 17295.6$ $J m o {l}^{-} 1 \approx 17.3$ $k J m o {l}^{-} 1$