14.7 g of solid sodium hydrogen carbonate (\(MM\) = 84.008 g mol\(^{-1}\)) was reacted with 120 mL of 1.50 mol L\(^{-1}\) hydrochloric acid solution (density 1.02 g mL\(^{-1}\)) in a calorimeter. The temperature of the solution before and after reaction was recorded. \( Assume that all \(\ce{CO2}\) produced is lost from the reaction solution and that the specific heat capacity of the reaction solution is 3.80 J K\(^{-1}\) g\(^{-1}\). What is the enthalpy of reaction, in kJ mol\(^{-1}\)? (5 marks) --- 15 WORK AREA LINES (style=lined) ---
\begin{array}{|c|c|}
\hline \begin{array}{c}
\textit {Initial temperature of } \\
\textit {hydrochloric acid solution } \\
\left({ }^{\circ} C\right)
\end{array} & \begin{array}{c}
\textit {Final temperature of } \\
\textit {reaction solution } \\
\left({ }^{\circ} C\right)
\end{array} \\
\hline 21.5 & 11.5 \\
\hline
\end{array}
\)
CHEMISTRY, M6 2020 HSC 36
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100.00 mL of 2.00 mol L¯1 \( \ce{HCl(aq)} \) was initially at a temperature of 22.5°C. The mass of this solution was 103 g.
10.0 g of solid `text{NaOH}` was added to the acid. The specific heat capacity of the resulting solution was 3.99 J g ¯1 K ¯1.
Assuming no energy loss to the environment, calculate the maximum temperature reached by the solution. (5 marks)
Use the following information in your calculations.
\begin{array} {ll}
\ce{NaOH(s) -> Na+(aq) + OH-(aq) & Δ$H=–44.5$\ \text{kJ mol}^{-1}} \\
\ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l) & Δ$H=–56.1$\ \text{kJ mol}^{-1}} \end{array}
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CHEMISTRY, M6 2020 HSC 25
Citric acid reacts with sodium hydroxide according to the following chemical equation:
\( \ce{C6H8O7(aq) + 3NaOH(aq) -> Na3C6H5O7(aq) + 3H2O(l)} \)
Various volumes of 1.0 mol L¯1 citric acid solution were mixed with 8.0 mL of a sodium hydroxide solution of unknown concentration and sufficient deionised water added to make the total volume of the resulting solution 14.0 mL. The change in temperature of each solution was measured.
The data are given in the table.
By graphing the data in the table and performing relevant calculations, determine the concentration of the sodium hydroxide solution. (7 marks)
--- 5 WORK AREA LINES (style=lined) ---
Show Answers Only
`text{n(citric acid)} = c xx V = (1.0) xx (2.4 xx 10^(−3)) = 2.4 xx 10^(−3)\ text{mol}`
`text{n(sodium hydroxide)} = 3 xx (2.4 xx 10^(−3)) = 7.2 xx 10^(−3)\ text{mol}`
`text{[sodium hydroxide]} = [7.2 xx 10^(−3)] / [8.0 xx 10^(−3)] = 0.90\ text{mol L}^(–1)`
Show Worked Solution
`text{n(citric acid)} = c xx V = (1.0) xx (2.4 xx 10^(−3)) = 2.4 xx 10^(−3)\ text{mol}`
`text{n(sodium hydroxide)} = 3 xx (2.4 xx 10^(−3)) = 7.2 xx 10^(−3)\ text{mol}`
`text{[sodium hydroxide]} = [7.2 xx 10^(−3)] / [8.0 xx 10^(−3)] = 0.90\ text{mol L}^(–1)`
♦ Mean mark 53%.
CHEMISTRY, M6 2022 HSC 26
Students conducted an experiment to determine `Delta H` for the reaction between sodium hydroxide and hydrochloric acid.
The data from one student are shown in the table below.
Assume that all the solutions have the same specific heat capacity as water.
- Calculate the heat energy released in this experiment. (2 marks)
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- A second student using the same procedure obtained `2.6 × 10^(3)\ text{J}` for the heat energy released in their experiment.
- Use this value to determine the enthalpy of neutralisation, `Delta H`, in `\text{kJ}\ text{mol}^(-1)`, for the reaction shown.
- \( \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)} \) (2 marks)
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CHEMISTRY, M6 2021 HSC 32
The molar enthalpies of neutralisation of three reactions are given.
Reaction 1:
\(\ce{HCl($aq$) + KOH($aq$) -> KCl($aq$) + H2O($l$)}\) \(\ce{Δ$H$}\) \(\pu{=-57.6 kJ mol-1}\)
Reaction 2:
\(\ce{HNO3($aq$) + KOH($aq$) -> KNO3($aq$) + H2O($l$)}\) \(\ce{Δ$H$}\) \(\pu{=-57.6 kJ mol-1}\)
Reaction 3:
\(\ce{HCN($aq$) + KOH($aq$) -> KCN($aq$) + H2O($l$)}\) \(\ce{Δ$H$}\) \(\pu{=-12.0 kJ mol-1}\)
Explain why the first two reactions have the same enthalpy value but the third reaction has a different value. (4 marks)
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