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

Calculate heat:

$\overline{){\mathbf{Q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{mC}}{\mathbf{\u2206}}{\mathbf{T}}}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{100}\mathbf{}\mathbf{g}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{184}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{g}\mathbf{\xb0}\mathbf{C}\mathbf{)}\mathbf{(}\mathbf{37}\mathbf{.}\mathbf{8}\mathbf{\xb0}\mathbf{C}\mathbf{-}\mathbf{21}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{\xb0}\mathbf{C}\mathbf{)}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{100}\mathbf{}\overline{)\mathbf{g}}\mathbf{)}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{184}\mathbf{}\mathbf{J}\mathbf{/}\overline{)\mathbf{g}}\overline{)\mathbf{\xb0}\mathbf{C}}\mathbf{)}\mathbf{(}\mathbf{16}\mathbf{.}\mathbf{2}\mathbf{}\overline{)\mathbf{\xb0}\mathbf{C}}\mathbf{)}\phantom{\rule{0ex}{0ex}}\mathbf{Q}\mathbf{}\mathbf{=}\mathbf{}\mathbf{6778}\mathbf{.}\mathbf{08}\mathbf{}\overline{)\mathbf{J}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{kJ}}{{\mathbf{10}}^{\mathbf{3}}\mathbf{}\overline{)\mathbf{J}}}$

**Q = 6.78 kJ**

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 ^{o}C to 37.8 ^{o}C.

Calculate ΔH (in kJ/mol NaOH) for the solution process

NaOH_{(s)} → Na^{+}_{(aq)} + OH^{-}_{(aq)}

Assume that the specific heat of the solution is the same as that of pure water.

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