Problem: When a 6.50-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (the following Figure), the temperature rises from 21.6 oC to 37.8 oC.Calculate ΔH (in kJ/mol NaOH) for the solution processNaOH(s) → Na+(aq) + OH-(aq)Assume that the specific heat of the solution is the same as that of pure water.

FREE Expert Solution

We are asked to calculate ΔH (in kJ/mol NaOH) for the solution process.

Calculate heat: 

$$\begin{gathered} \boxed{\mathbf{Q}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{mC}\mathbf{∆}\mathbf{T}} \\ \mathbf{Q}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{(}\mathbf{100}\mathbf{ }\mathbf{g}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{184}\mathbf{ }\mathbf{J}\mathbf{/}\mathbf{g}\mathbf{°}\mathbf{C}\mathbf{)}\mathbf{(}\mathbf{37}\mathbf{.}\mathbf{8}\mathbf{°}\mathbf{C}\mathbf{-}\mathbf{21}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{°}\mathbf{C}\mathbf{)} \\ \mathbf{Q}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{(}\mathbf{100}\mathbf{ }\cancel{\mathbf{g}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{184}\mathbf{ }\mathbf{J}\mathbf{/}\cancel{\mathbf{g}}\cancel{\mathbf{°}\mathbf{C}}\mathbf{)}\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{2}\mathbf{ }\cancel{\mathbf{°}\mathbf{C}}\mathbf{)} \\ \mathbf{Q}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{6778}\mathbf{.}\mathbf{08}\mathbf{ }\cancel{\mathbf{J}}\mathbf{ }\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{kJ}}{{\mathbf{10}}^{\mathbf{3}}\mathbf{ }\cancel{\mathbf{J}}} \end{gathered}$$

Q = 6.78 kJ