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Looking for more practice with Module 4: Drivers of Reactions? Well, you came to the right place! We give you 10 questions of varying difficulties and their solutions to test your knowledge!
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Practice is the best way to develop your Chemistry skills. Below are some Module 4: Drivers of Reactions practice questions to prepare you for your Chemistry yearly exams.
Below are 9 questions to test your knowledge of drivers of reaction. You can find the worked solutions here, at the bottom of the page.
Identify the sign of \( q_{systems}\) and \( q_{surroundings} \) for the following situations.
| Situation | Sign of \( q_{systems} \) | Sign of \( q_{surroundings} \) |
| A metal spoon (system) feels hot after being placed in a cup of hot tea for a few minutes. | ||
| A can of drink (system) at room temperature is placed in a fridge. | ||
| An apple pie (system) is baked in an oven. | ||
| When aqueous solutions of \( HCl \ (1.0\ M) \) and \( NaOH \ (1.0 \ M) \) are mixed, the solution (surroundings) becomes hot. | ||
| Ice cream (system) melts when kept at room temperature. |
Calculate the amount of energy produced or absorbed by the following processes:
(a) Raising the temperature of \( 20.0 \ g \) of water from \( 15.0 °C \) to \( 75.0 °C \) . (1 mark)
(b) Heating \( 500.0 \ g \) of iron from \( 303 \ K \) to \( 423 \ K \) given that the specific heat capacity of iron is \( 0.444 \times 10^{3} \ J \ kg^{-1} K^{-1}\). (1 mark)
(c) Cooling \( 25 mL \) of ethanol from \( 25.0 °C \) to \( 6.0 °C \). Ethanol has a density of \( 0.785 \ g \ cm^{−3} \) and a specific heat capacity of \( 2.46 \times 10^{3} \ J \ kg^{-1} \ K^{-1}\). (2 marks)
(a) Calculate the heat energy that is released when a \( 23.0 \ g \) sample of ethanol is cooled from \( 45.0 °C \) to room temperature \( (25.0 °C) \) . The specific heat capacity of ethanol is \( 2.46 \times 10^{3} \ J \ kg^{-1} \ K^{-1}\). b
(b) \(4.2 g \) of propan-1-ol \( (C_{3}H_{7}OH) \) was combusted to heat \(750 \ g \) of water to \(80.0 °C \). If the enthalpy of combustion of propan-1-ol is \( -2021\ kJ \ mol^{-1}\), calculate the starting temperature of the water. (2 marks)
(c) A student performed an experiment to determine the molar enthalpy of combustion of butan‑1‑ol. The enthalpy value determined in the experiment was \( -1740 \ kJ \ mol^{-1}\), however the published value for the molar enthalpy of combustion of butan-1-ol is \(-2671 \ kJ \ mol^{-1} \). Provide a reason for the discrepancy between these values, and suggest a way to improve the accuracy of the experiment. (2 marks)
A coffee cup calorimeter was used to measure the enthalpy of solution \( (Δ_{sol}H^{o}) \) of ammonium chloride.
(a) Write the net ionic equation for the dissolution of ammonium chloride. (1 mark)
(b) A \( 10.0 \ g \) sample of ammonium chloride was dissolved in \( 100 \ g \) of water, causing the temperature to decrease from \( 30.0 °C \) to \( 26.0 °C \) . Calculate the enthalpy of solution of ammonium chloride. (2 marks)
Calculate the standard enthalpy of reaction \( (ΔH^{o}) \) for the following reaction:
\( NO_{(g)} + \frac{1}{2} O_{2(g)} \ → \ NO_{2(g)} \)
Given the following thermochemical data:
(1) \( 2O_{3(g)} → 3O_{2(g)} \)
\( ΔH^{o} \ = \ −285 \ kJ \ mol^{−1} \)
(2) \( NO_{2(g)} + O_{2(g)} → NO_{(g)} + O_{3(g)} \)
\( ΔH^{o} = 199 \ kJ \ mol^{−1} \)
The following table shows the bond enthalpy values for some common chemical bonds.
| Bond | Bond enthalpy \( (kJ \ mol^{-1}) \) |
| C-H | 413 |
| C-C | 348 |
| O=O | 495 |
| O-O | 146 |
| C=O | 358 |
| C=O | 799 |
| C≡O | 1072 |
| O-H | 463 |
(a) Explain, with reference to bond enthalpy, why some reactions are endothermic while others are exothermic. (1 mark)
(b) Use the bond enthalpy values in the table above to calculate the enthalpy change for the following reaction: (2 marks)
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(c) Challenge: Use the bond enthalpy values above to calculate the enthalpy of combustion of propane \( (C_{3}H_{8(g)}) \). The enthalpy of vaporisation of water \( (H_{2}O_{(l)} \ → \ H_{2}O_{(g)} ) \) is \( 41 \ kJ \ mol^{-1}\). (2 marks)
For the following reactions, predict the sign of the entropy change \( (ΔS) \) .
(a) \( CaCO_{3(s)} \ → \ CaO_{(s)} \ + \ CO_{2(g)} \)
(b) \( 2HCl_{(aq)} \ + \ Zn_{(s)} \ → \ ZnCl_{2(aq)} \ + \ H_{2(g)} \)
(c) \( 14O_{2(g)} \ + \ 3NH_{4}NO_{3(s)} \ + \ C_{10}H_{22(l)} \ → \ 3N_{2(g)} + 17H_{2}O_{(g)} \ + \ 10CO_{2(g)} \)
(d) \( N_{2(g)} \ + \ 3H_{2(g)} \ → \ 2NH_{3(g)} \)
A 2.2 g sample of \( CaCO_{3(s)} \) was decomposed to \( CaO_{(s)} \) and \( CO_{2(g)} \). The standard entropy values for the reactants and products are shown in the table below.
| Substance | Standard entropy \( S° \ (J \ mol^{-1} \ KY^{-1} ) \) |
| \( CaCO_{3(s)} \) | \( 92.88 \) |
| \( CaO_{(s)} \) | \( 39.75 \) |
| \( CO_{2(g)} \) | \( 213.6 \) |
(a) Write a chemical equation for this process. (1 mark)
(b) Calculate the entropy change for this reaction. (1 mark)
(c) \( 7.942 \ kJ \) of heat energy \( (q) \) was absorbed by the decomposition reaction. Calculate the molar enthalpy of the reaction. (2 marks)
(d) Determine whether the reaction is spontaneous or non-spontaneous at room temperature. (2 marks)
Ammonia gas is produced industrially through the reaction of nitrogen and hydrogen gases.
The standard enthalpies of formation \( (ΔH_{f^{o}}) \) and standard entropies \( (S^{o}) \) for the reactants and products are shown in the table below.
| Compound | \( ΔH_{f^{o}} \ (kJ \ mol^{-1}) \) | \( S^{o} \ (J \ K^{-1} \ mol{-1}) \) |
| \( N_{2(g)} \) | \( 0 \) | \( 191.61 \) |
| \( H_{2(g)} \) | \( 0 \) | \( 130.68 \) |
| \( NH_{3(g)} \) | \( -46.11 \) | \( 192.45 \) |
(a) Write a balanced chemical equation for the reaction. (1 mark)
(b) Calculate the enthalpy and entropy change for the reaction. (4 marks)
(c) Using the values calculated in part (b), calculate the temperature range for which the reaction is spontaneous. (2 marks)
The enthalpy of fusion of water is the enthalpy change associated with a change in state from a solid to a liquid: \( H_{2}O_{(s)} \ → \ H_{2}O_{(l)} \).
The standard enthalpy of formation for \( H_{2}O_{(s)} \) is \( -291.83 \ kJ \ mol^{-1} \) and the standard enthalpy of formation for \( H_{2}O_{(l)} \) is \( -285.8 \ kJ \ mol^{-1}\).
(a) Calculate the enthalpy of fusion of water. (1 mark)
(b) Calculate the entropy of fusion at the melting point of \( 0 °C \). Note: the Gibbs free energy at a phase transition is \(0\). (2 marks)
| Situation | Sign of \( q_{systems} \) | Sign of \( q_{surroundings} \) |
| A metal spoon (system) feels hot after being placed in a cup of hot tea for a few minutes. | Positive \( (+) \) | Negative \((-) \) |
| A can of drink (system) at room temperature is placed in a fridge. | Negative \((-) \) | Positive \( (+) \) |
| An apple pie (system) is baked in an oven. | Positive \( (+) \) | Negative \((-) \) |
| When aqueous solutions of \( HCl \ (1.0 M) \) and \( NaOH \ (1.0 M) \) are mixed, the solution (surroundings) becomes hot. | Negative \((-) \) | Positive \( (+) \) |
| Ice cream (system) melts when kept at room temperature. | Positive \( (+) \) | Negative \((-) \) |
(a)
| \( q \ = \ mcΔT \ = \ 0.0200 \ \times \ 4.18 \ \times \ 10^{3} \ \times \ (75.0 \ – \ 15.0) \ = \ 5016 \ = \ 5020 \ J \) produced (3 sig figs) |
(b)
| \( q \ = \ mcΔT \ = \ 0.500 \ \times \ 0.444 \ \times \ 10^{3} \ \times \ (423 \ – \ 303) \ = \ 26640 \ = \ 26600 \ J \) produced (3 sig figs) |
(c)
| Mass of ethanol \( = \ d \ \times \ V \ = \ 0.785 \ \times \ 25 \ = \ 19.625 \ g. \) \( q \ = \ mcΔT \ = \ 0.019625 \ \times \ 2.46 \ \times \ 10^{3} \ \times \ (6.0 \ – \ 25.0) \ = \ -917.2725 \ = \ 920 \ J \) absorbed (2 sig figs) |
(a)
| \( q \ = \ mcΔT \ = \ 23 \ \times \ 10^{-3} \ \times \ 2.46 \ \times \ 10^{3} \ \times \ (25.0 \ – \ 45.0) \ = \ -1131.6 \ = \ -1130 \ J \) (3 sig figs) |
(b)
| \begin{align*} ΔH \ &= \ \frac{-q}{n} \ → q \ = \ -ΔH \ \times \ n \\ n(propan-1-ol) \ &= \ \frac{4.2}{60.094} \ = \ 0.0698905 \ \text{mol} \\ q \ &= \ 2021 \ \times \ 0.0698905 \ = \ 141.2487 \ kJ \ = \ 141248.7104 \ J \\ q \ &= \ mcΔT \ = \ mc(T_{Final} \ – \ T_{Initial}) \ → \ T_{initial} \ = \ T_{final} \ – \ \frac{q}{mc} \ = \ 80.0 °C \ – \ \big{(} \frac{141248.7104}{(0.750 \ \times \ 4.18 \ \times \ 10^{3}} \big{)} \ = \ 80.0 \ – \ 45.0676 \ = \ 35°C \ \text{(2 sig figs)} \end{align*} |
(c)
| The experiment makes an assumption that all of the heat energy released in the combustion reaction is absorbed by the water. In reality, there is heat loss to the surroundings which accounts for the lower experimental value for molar enthalpy of combustion. Insulating the calorimeter will prevent heat loss to the surroundings and improve the accuracy of the experiment. |
(a)
| \( NH_{4}Cl_{(s)} \ → \ NH_{4^{+}(aq)} \ + \ Cl- _{(aq)} \) |
(b)
| \begin{align*} n \ &= \ \frac{m}{MM} \ = \ \frac{10.0}{53.492} \ = \ 0.186944 \ mol \\ q \ &= \ mcΔT \ = 0.110 \ \times \ 4.18 \ \times \ 103 \ \times \ (26.0 \ – \ 30.0) \ = \ -1839.2 \ J \ = \ -1.8392 \ kJ \\ ΔH & = \ \frac{-q}{n} \ = \ \frac{1.8392}{0.186944} \ = \ 9.8 \ kJ \ mol^{-1} \ \text{(2 sig figs)} \end{align*} |
(1)
| Reverse, divide by 2: \( \frac{3}{2O_{2(g)}} \ → \ O_{3(g)} \) \( ΔH° \ = \ 142.5 \ kJ \ mol^{−1} \) |
(2)
| Reverse: \( NO_{(g)} \ + \ O{3(g)} \ → \ O_{2(g)} \ + \ NO_{2(g)} \)
\( ΔH° \ = \ -199 \ kJ \ mol^{−1} \)
Adding the equations together gives the target equation: \( NO_{(g)} \ + \ \frac{1}{2} O_{2(g)} → NO_{2(g)} \)\( ΔH° \ = \ −57 \ kJ \ mol^{−1} \) |
(a) Bond enthalpy is the average energy required to break one mole of a bond when a substance is in a gaseous state. This is also the same amount of energy that is released when one mole of a bond forms between atoms. Both bond breaking and bond-forming processes occur in a chemical reaction.
If the energy required to break bonds in a chemical reaction is greater than the energy that is released when bonds form, the overall reaction will be endothermic. Conversely, if the energy released during bond formation is greater than the energy required to break bonds, the reaction will be exothermic.
(b)
| \begin{align*} ΔH &= ΣΔH_{\text{reactant bonds}} \ − \ Σ ΔH_{\text{product bonds}}\\ &= \big{(} (2 \ \times \ C≡O) \ + \ (1 \ \times \ O=O) \big{)} \ – \ (4 \ \times \ C=O) \\ &= \big{(} (2 \ \times \ 1072) \ + \ (495) \big{)} \ – \ (4 \ \times \ 799) \\ &= \ -557 \ kJ \ mol^{-1} \end{align*} |
(c)
| \( C_{3}H_{8(g)} \ + \ 5O_{2(g)} \ → \ 3CO_{2(g)} \ + \ 4H_{2}O_{(g)} \) image \begin{align*}
The enthalpy of combustion, however, requires that water is in the liquid state: Target equation: \( C_{3}H_{8(g)} \ + \ 5O_{2(g)} \ → \ 3CO_{2(g)} \ + \ 4H_{2}O_{(l)} \) In order to find the enthalpy of combustion for propane, the enthalpy of vaporisation data must also be considered. (1) \(C_{3}H_{8(g)} \ + \ 5O_{2(g)} \ → \ 3CO_{2(g)} \ + \ 4H_{2}O_{(g)} \) \( ΔH \ = \ -2023 \ kJ \ mol^{-1} \)
(2) \( H_{2}O_{(l)} \ → \ H_{2}O_{(g)} \) \( ΔH \ = \ 41\ kJ \ mol^{-1} \)
To obtain the target equation, (2) must be reversed: (1) \( C_{3}H_{8(g)} \ + \ 5O_{2(g)} \ → \ 3CO_{2(g)} \ + \ 4H_{2}O_{(g)} \) \( ΔH \ = \ -2023 \ kJ \ mol^{-1} \)
(2) \( H_{2}O_{(g)} → H_{2}O_{(l)} \) \( ΔH \ = \ -41 \ kJ \ mol^{-1} \)Adding these equations gives \( ΔH \ = \ -2064 \ kJ \ mol^{-1}\) |
(a) Positive – A solid is decomposed to produce a gaseous product.
(b) Positive – A solid substance reacts to produce a gaseous product.
(c) Positive – There is an increase in the number of moles of substances in the gas state.
(d) Negative – Four moles of gaseous reactants combine to form two moles of a gaseous product.
(a)
| \( CaCO_{3(s)} \ → \ CaO_{(s)} \ + \ CO_{2(g)} \) |
(b)
| \begin{align*} ΔS^{o} \ &= \ ΣS^{o} \ _{products} – ΣS^{o} \ _{reactants} \\ &= \ (213.6 \ + \ 39.75) \ – \ 92.88 \ = \ 160.47 \ J \ mol^{-1}K^{-1}\\ n(CaCO_{3}) &= \frac{m}{MM} \ = \ \frac{2.2}{100.09} \\ &= \ 0.02198 \ mol \end{align*}Entropy change for this reaction \( = \ 160.47 \ J \ mol^{-1} \ K^{-1} \ \times \ 0.02198 \ mol \ = \ 3.5 \ J \ K^{-1}\) (2 sig figs) |
(c)
| \begin{align*} ΔH \ &= \frac{-q}{n} \ = \ \frac{7.942}{0.02198} \ = \ 361.3 \\ &= \ 360 \ kJ \ mol^{-1} \text{(2 sig figs)} \end{align*} |
(d)
| \begin{align*} \text{Room temperature} \ &= \ 298.15 \ K.\\ ΔG^{o} \ &= \ ΔH^{o} \ – \ TΔS^{o} \\ ΔG \ &= \ 361.3 \ – \ 298.15 \ \times \ 0.16047 \ = \ 313.48076 \ = \ 310 \ kJ \ mol^{-1} \\ ΔG \ &> \ 0, \ \text{therefore nonspontaneous} \end{align*} |
(a)
| \( N_{2(g)} \ + \ 3H_{2(g)} \ → \ 2NH_{3(g)} \) |
(b)
| \begin{align*} ΔH^{o} \ &= \ ΣΔH_{f^{o}products} \ − \ Σ ΔH_{f^{o}reactants} \\ &= \ (2 \ \times \ -46.11) \ – \ (0) \ = \ -92.22 \ kJ \ mol^{-1} \\ ΔS^{o} \ &= \ ΣS^{o} \ _{products} \ – \ ΣS^{o} \ _{reactants} \\ &= \ (2 \times 192.45) \ – \ \big{(} (3 \ \times \ 130.68) \ + \ (191. 61) \big{)} \ = \ -198.75 \ J \ K^{-1} \ mol^{-1} \\ &= \ -0.19875 \ kJ \ K^{-1} \ mol^{-1} \end{align*} |
(c)
| \begin{align*} &ΔG^{o} \ = \ ΔH^{o} \ – \ TΔS^{o}. \text{For spontaneous reactions}, ΔG \ < \ 0 \\ &ΔH^{o} \ – \ TΔS^{o} \ < 0\\ Therefore, \ T \ &< \ \frac{92.22}{0.19875}; \ T \ < \ 464 \ K \end{align*} |
a)
| \begin{align*} ΔH \ &= \ ΣΔH_{f^{o}products} \ − \ Σ ΔHf_{f^{o}reactants} \\ &= \ -285.8 \ + \ 291.83 \\ &= \ 6.03 kJ \ mol^{-1} \end{align*} |
(b)
| \begin{align*} \text{At melting point}, \ ΔG^{o} \ &= \ 0 \ → \ ΔH^{o} \ – \ TΔS^{o} \ = \ 0 \ → \ ΔS^{o} \ = \ \frac{ΔH^{o}}{T} \\ ΔS^{o} &= \ \frac{6.03}{273.15} \ = \ 0.02207578 \ = \ 22 \ J \ K^{-1} \ mol^{-1} \end{align*} |
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