In this experiment you will determine the enthalpy change, $$\(\Delta H_{r}\)$$, for the reaction shown. $$\[ \mathrm{NaOH}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{NaHCO}_{3}(\mathrm{aq}) \quad \text { enthalpy change }=\Delta H_{\mathrm{r}} \]$$ You will react each of sodium hydroxide and sodium hydrogencarbonate with excess dilute sulfuric acid. You will determine the enthalpy change for each reaction, then use Hess's law to calculate $$\(\Delta H_{r}\)$$. The enthalpy change when one mole of sodium hydrogencarbonate dissolves in water is $$\(\Delta H_{3}\)$$. $$\[ \mathrm{NaHCO}_{3}(\mathrm{~s})+\mathrm{aq} \rightarrow \mathrm{NaHCO}_{3}(\mathrm{aq}) \quad \text { enthalpy change }=\Delta H_{3} \]$$ Using the symbols $$\(\Delta H_{1}, \Delta H_{2}\)$$ and $$\(\Delta H_{3}\)$$ in your answer, use Hess's law to deduce an expression for $$\(\Delta H_{r}\)$$. $$\[ \mathrm{NaOH}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g}) \rightarrow \mathrm{NaHCO}_{3}(\mathrm{aq}) \quad \text { enthalpy change }=\Delta H_{\mathrm{r}} \]$$

Answer:

Knowledge points:

Solution: