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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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`T_(f i nal) = 72.1text{°C  (to 1 d.p.)}`

Show Worked Solution

\( \text{Dissolution of}\ \ce{NaOH(s):} \)

\(\ce{NaOH(s) → Na+ (aq) + OH– (aq)}\)

`text{n(NaOH)} = text{m} / text{MM} = [10.0] / [22.99 + 16.00 + 1.008] = 0.250\ text{mol}`

`q_1 = -ΔH xx text{n} = 44.5 xx 0.250 = −11.125\ text{kJ}`
 

\( \text{Reaction between}\ \ce{NaOH(aq)}\ \text{and}\ \ce{HCl(aq):}\)

\(\ce{HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)}\)

`text{n(HCl)} = text{c} xx text{V} = 2.00 xx 0.100 = 0.200\ text{mol}`

`=>\ text{HCl = limiting reagent}`
 

`text{n(H}_2 text{O) formed} = 0.200\ text{mol}`

`q_2 = -ΔH xx text{n} = 56.1 xx 0.200 = 11.22\ text{kJ}`

`q_(t otal) = q_1 + q_2 = 11.125 + 11.22 = 22.345\ text{kJ} = 22\ 345\ text{J}`
 

`q_(t otal)` `= mcΔT`  
`22\ 345` `= (103 + 10.0) xx 3.99 xx (T_(f i nal) − T_(i nitial))`  
  `= (103 + 10.0) xx 3.99 xx (T_(f i nal) − 22.5)`  
`T_(f i nal)` `=(22\ 345)/((103 + 10.0) xx 3.99)+22.5`  
  `= 72.1text{°C  (to 1 d.p.)}`  

♦ Mean mark 49%.

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)
 

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`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.

  1. Calculate the heat energy released in this experiment.   (2 marks)
  1. A second student using the same procedure obtained `2.6 × 10^(3)\ text{J}` for the heat energy released in their experiment.
  2. Use this value to determine the enthalpy of neutralisation, `Delta H`, in `\text{kJ}\ text{mol}^(-1)`, for the reaction shown.
  3.      \( \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)} \)   (2 marks)
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  1. `2800  text{J}`
  2. `-52\ text{kJ mol}^-1`
Show Worked Solution

a.   `T_(text{avg initial}) = (21.0 + 21.2) / 2 = 21.1 text{°C}`

`q` `= mc DeltaT`
  `= (100.7 + 102.0) (4.18) (24.4 − 21.1)`
  `=2796.04…`
  `= 2800  text{J (to 2 s.f.)}`

 
Therefore, 2800 J of heat energy is released in this experiment (assuming no energy loss).
 

b.   `q = 2600\ text{J}`

`text{n} = 0.1 xx 0.5 = 0.05\ text{mol}`

`DeltaH` `= -q/text{n}`
  `= -2600 / 0.05`
  `= -52\ 000\ text{J mol}^-1`
  `= -52\ text{kJ mol}^-1`

♦ Mean mark (b) 43%.

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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→ Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.

→ Both reactions have the same net ionic equation:

`text{H}^+ (aq) + text{OH}^-  (aq) →  text{H}_2 text{O} (l)`

→ Therefore, the enthalpy values obtained are the same for both reactions.

→ In reaction 3, `text{HCN}` is a weak acid that only partially ionises in an equilibrium reaction with water.

`text{HCN} (aq) + text{H}_2 text{O} (l)  ⇋  text{CN}^-  (aq) + text{H}_3 text{O}^+ (aq).`

→ As the reaction continues, `text{HCN}` will further ionise as the equilibrium shifts to the right.

→ The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.

Show Worked Solution

→ Reaction 1 and reaction 2 are both neutralisation reactions between strong acids and strong bases. These reactions completely ionise in solution when added to water.

→ Both reactions have the same net ionic equation:

`text{H}^+ (aq) + text{OH}^-  (aq) →  text{H}_2 text{O} (l)`

→ Therefore, the enthalpy values obtained are the same for both reactions.

→ In reaction 3, `text{HCN}` is a weak acid that only partially ionises in an equilibrium reaction with water.

`text{HCN} (aq) + text{H}_2 text{O} (l)  ⇋  text{CN}^-  (aq) + text{H}_3 text{O}^+ (aq).`

→ As the reaction continues, `text{HCN}` will further ionise as the equilibrium shifts to the right.

→ The bond-breaking is an endothermic process and thus will consume energy to break the bonds. As a result, the overall reaction is less exothermic than reaction 1 and reaction 2.


♦ Mean mark 44%.