Observe is one of the best ways to develop your Chemistry abilities. Under are some Module 4: Drivers of Reactions follow questions to organize you to your Chemistry yearly exams.
12 months 11 Chemistry Mod 4: Drivers of Reactions Observe Questions
Under are 9 questions to check your data of drivers of response. Yow will discover the labored options right here, on the backside of the web page.
Questions
Query 1 – Vitality (5 marks)
Establish the signal of ( q_{techniques}) and ( q_{environment} ) for the next conditions.
| State of affairs | Signal of ( q_{techniques} ) | Signal of ( q_{environment} ) |
| A steel spoon (system) feels sizzling after being positioned in a cup of sizzling tea for a couple of minutes. | ||
| A can of drink (system) at room temperature is positioned in a fridge. | ||
| An apple pie (system) is baked in an oven. | ||
| When aqueous options of ( HCl (1.0 M) ) and ( NaOH (1.0 M) ) are combined, the answer (environment) turns into sizzling. | ||
| Ice cream (system) melts when stored at room temperature. |
Query 2 – Warmth power
Calculate the quantity of power produced or absorbed by the next processes:
(a) Elevating 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 Okay ) to ( 423 Okay ) provided that the particular warmth capability of iron is ( 0.444 instances 10^{3} J kg^{-1} Okay^{-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 particular warmth capability of ( 2.46 instances 10^{3} J kg^{-1} Okay^{-1}). (2 marks)
Query 3 – Enthalpy of combustion
(a) Calculate the warmth power that’s launched when a ( 23.0 g ) pattern of ethanol is cooled from ( 45.0 °C ) to room temperature ( (25.0 °C) ) . The precise warmth capability of ethanol is ( 2.46 instances 10^{3} J kg^{-1} Okay^{-1}). b
(b) (4.2 g ) of propan-1-ol ( (C_{3}H_{7}OH) ) was combusted to warmth (750 g ) of water to (80.0 °C ). If the enthalpy of combustion of propan-1-ol is ( -2021 kJ mol^{-1}), calculate the beginning temperature of the water. (2 marks)
(c) A pupil carried out an experiment to find out the molar enthalpy of combustion of butan‑1‑ol. The enthalpy worth decided within the experiment was ( -1740 kJ mol^{-1}), nevertheless the revealed worth for the molar enthalpy of combustion of butan-1-ol is (-2671 kJ mol^{-1} ). Present a purpose for the discrepancy between these values, and counsel a method to enhance the accuracy of the experiment. (2 marks)
Query 4 – Enthalpy of answer
A espresso cup calorimeter was used to measure the enthalpy of answer ( (Δ_{sol}H^{o}) ) of ammonium chloride.
(a) Write the web ionic equation for the dissolution of ammonium chloride. (1 mark)
(b) A ( 10.0 g ) pattern of ammonium chloride was dissolved in ( 100 g ) of water, inflicting the temperature to lower from ( 30.0 °C ) to ( 26.0 °C ) . Calculate the enthalpy of answer of ammonium chloride. (2 marks)
Query 5 – Enthalpy and Hess’s Legislation (2 marks)
Calculate the usual enthalpy of response ( (ΔH^{o}) ) for the next response:
( NO_{(g)} + frac{1}{2} O_{2(g)} → NO_{2(g)} )
Given the next thermochemical information:
(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} )
Query 6 – Bond energies
The next desk reveals the bond enthalpy values for some frequent 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) Clarify, as regards to bond enthalpy, why some reactions are endothermic whereas others are exothermic. (1 mark)
(b) Use the bond enthalpy values within the desk above to calculate the enthalpy change for the next response: (2 marks)
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(c) Problem: 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)
Query 7 – Entropy (4 marks)
For the next reactions, predict the signal 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)} )
Query 8 – Entropy, enthalpy and Gibbs free power
A 2.2 g pattern of ( CaCO_{3(s)} ) was decomposed to ( CaO_{(s)} ) and ( CO_{2(g)} ). The usual entropy values for the reactants and merchandise are proven within the desk beneath.
| Substance | Customary 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 course of. (1 mark)
(b) Calculate the entropy change for this response. (1 mark)
(c) ( 7.942 kJ ) of warmth power ( (q) ) was absorbed by the decomposition response. Calculate the molar enthalpy of the response. (2 marks)
(d) Decide whether or not the response is spontaneous or non-spontaneous at room temperature. (2 marks)
Query 9 – Gibbs free power
Ammonia gasoline is produced industrially by the response of nitrogen and hydrogen gases.
The usual enthalpies of formation ( (ΔH_{f^{o}}) ) and commonplace entropies ( (S^{o}) ) for the reactants and merchandise are proven within the desk beneath.
| Compound | ( ΔH_{f^{o}} (kJ mol^{-1}) ) | ( S^{o} (J Okay^{-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 response. (1 mark)
(b) Calculate the enthalpy and entropy change for the response. (4 marks)
(c) Utilizing the values calculated partly (b), calculate the temperature vary for which the response is spontaneous. (2 marks)
Query 10 – Problem
The enthalpy of fusion of water is the enthalpy change related to a change in state from a stable to a liquid: ( H_{2}O_{(s)} → H_{2}O_{(l)} ).
The usual enthalpy of formation for ( H_{2}O_{(s)} ) is ( -291.83 kJ mol^{-1} ) and the usual 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 on the melting level of ( 0 °C ). Word: the Gibbs free power at a section transition is (0). (2 marks)
Options
Query 1
| State of affairs | Signal of ( q_{techniques} ) | Signal of ( q_{environment} ) |
| A steel spoon (system) feels sizzling after being positioned in a cup of sizzling tea for a couple of minutes. | Constructive ( (+) ) | Damaging ((-) ) |
| A can of drink (system) at room temperature is positioned in a fridge. | Damaging ((-) ) | Constructive ( (+) ) |
| An apple pie (system) is baked in an oven. | Constructive ( (+) ) | Damaging ((-) ) |
| When aqueous options of ( HCl (1.0 M) ) and ( NaOH (1.0 M) ) are combined, the answer (environment) turns into sizzling. | Damaging ((-) ) | Constructive ( (+) ) |
| Ice cream (system) melts when stored at room temperature. | Constructive ( (+) ) | Damaging ((-) ) |
Query 2
(a)
| ( q = mcΔT = 0.0200 instances 4.18 instances 10^{3} instances (75.0 – 15.0) = 5016 = 5020 J ) produced (3 sig figs) |
(b)
| ( q = mcΔT = 0.500 instances 0.444 instances 10^{3} instances (423 – 303) = 26640 = 26600 J ) produced (3 sig figs) |
(c)
| Mass of ethanol ( = d instances V = 0.785 instances 25 = 19.625 g. )
( q = mcΔT = 0.019625 instances 2.46 instances 10^{3} instances (6.0 – 25.0) = -917.2725 = 920 J ) absorbed (2 sig figs) |
Query 3
(a)
| ( q = mcΔT = 23 instances 10^{-3} instances 2.46 instances 10^{3} instances (25.0 – 45.0) = -1131.6 = -1130 J ) (3 sig figs) |
(b)
| start{align*} ΔH &= frac{-q}{n} → q = -ΔH instances n n(propan-1-ol) &= frac{4.2}{60.094} = 0.0698905 textual content{mol} q &= 2021 instances 0.0698905 = 141.2487 kJ = 141248.7104 J q &= mcΔT = mc(T_{Last} – T_{Preliminary}) → T_{preliminary} = T_{remaining} – frac{q}{mc} = 80.0 °C – huge{(} frac{141248.7104}{(0.750 instances 4.18 instances 10^{3}} huge{)} = 80.0 – 45.0676 = 35°C textual content{(2 sig figs)} finish{align*} |
(c)
| The experiment makes an assumption that the entire warmth power launched within the combustion response is absorbed by the water. In actuality, there’s warmth loss to the environment which accounts for the decrease experimental worth for molar enthalpy of combustion. Insulating the calorimeter will stop warmth loss to the environment and enhance the accuracy of the experiment. |
Query 4
(a)
| ( NH_{4}Cl_{(s)} → NH_{4^{+}(aq)} + Cl- _{(aq)} ) |
(b)
| start{align*} n &= frac{m}{MM} = frac{10.0}{53.492} = 0.186944 mol q &= mcΔT = 0.110 instances 4.18 instances 103 instances (26.0 – 30.0) = -1839.2 J = -1.8392 kJ ΔH & = frac{-q}{n} = frac{1.8392}{0.186944} = 9.8 kJ mol^{-1} textual content{(2 sig figs)} finish{align*} |
Query 5
(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} )
Including the equations collectively offers the goal equation: ( NO_{(g)} + frac{1}{2} O_{2(g)} → NO_{2(g)} ) ( ΔH° = −57 kJ mol^{−1} ) |
Query 6
(a) Bond enthalpy is the common power required to interrupt one mole of a bond when a substance is in a gaseous state. That is additionally the identical quantity of power that’s launched when one mole of a bond kinds between atoms. Each bond breaking and bond-forming processes happen in a chemical response.
If the power required to interrupt bonds in a chemical response is larger than the power that’s launched when bonds kind, the general response shall be endothermic. Conversely, if the power launched throughout bond formation is larger than the power required to interrupt bonds, the response shall be exothermic.
(b)
| start{align*} ΔH &= ΣΔH_{textual content{reactant bonds}} − Σ ΔH_{textual content{product bonds}} &= huge{(} (2 instances C≡O) + (1 instances O=O) huge{)} – (4 instances C=O) &= huge{(} (2 instances 1072) + (495) huge{)} – (4 instances 799) &= -557 kJ mol^{-1} finish{align*} |
(c)
| ( C_{3}H_{8(g)} + 5O_{2(g)} → 3CO_{2(g)} + 4H_{2}O_{(g)} )
picture start{align*}
The enthalpy of combustion, nevertheless, requires that water is within the liquid state: Goal equation: ( C_{3}H_{8(g)} + 5O_{2(g)} → 3CO_{2(g)} + 4H_{2}O_{(l)} ) So as to discover the enthalpy of combustion for propane, the enthalpy of vaporisation information should even be thought-about. (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 acquire the goal equation, (2) should 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} ) Including these equations offers ( ΔH = -2064 kJ mol^{-1}) |
Query 7
(a) Constructive – A stable is decomposed to provide a gaseous product.
(b) Constructive – A stable substance reacts to provide a gaseous product.
(c) Constructive – There is a rise within the variety of moles of gear within the gasoline state.
(d) Damaging – 4 moles of gaseous reactants mix to kind two moles of a gaseous product.
Query 8
(a)
| ( CaCO_{3(s)} → CaO_{(s)} + CO_{2(g)} ) |
(b)
| start{align*} ΔS^{o} &= ΣS^{o} _{merchandise} – ΣS^{o} _{reactants} &= (213.6 + 39.75) – 92.88 = 160.47 J mol^{-1}Okay^{-1} n(CaCO_{3}) &= frac{m}{MM} = frac{2.2}{100.09} &= 0.02198 mol finish{align*}Entropy change for this response ( = 160.47 J mol^{-1} Okay^{-1} instances 0.02198 mol = 3.5 J Okay^{-1}) (2 sig figs) |
(c)
| start{align*} ΔH &= frac{-q}{n} = frac{7.942}{0.02198} = 361.3 &= 360 kJ mol^{-1} textual content{(2 sig figs)} finish{align*} |
(d)
| start{align*} textual content{Room temperature} &= 298.15 Okay. ΔG^{o} &= ΔH^{o} – TΔS^{o} ΔG &= 361.3 – 298.15 instances 0.16047 = 313.48076 = 310 kJ mol^{-1} ΔG &> 0, textual content{subsequently nonspontaneous} finish{align*} |
Query 9
(a)
| ( N_{2(g)} + 3H_{2(g)} → 2NH_{3(g)} ) |
(b)
| start{align*} ΔH^{o} &= ΣΔH_{f^{o}merchandise} − Σ ΔH_{f^{o}reactants} &= (2 instances -46.11) – (0) = -92.22 kJ mol^{-1} ΔS^{o} &= ΣS^{o} _{merchandise} – ΣS^{o} _{reactants} &= (2 instances 192.45) – huge{(} (3 instances 130.68) + (191. 61) huge{)} = -198.75 J Okay^{-1} mol^{-1} &= -0.19875 kJ Okay^{-1} mol^{-1} finish{align*} |
(c)
| start{align*} &ΔG^{o} = ΔH^{o} – TΔS^{o}. textual content{For spontaneous reactions}, ΔG < 0 &ΔH^{o} – TΔS^{o} < 0 Subsequently, T &< frac{92.22}{0.19875}; T < 464 Okay finish{align*} |
Query 10
a)
| start{align*} ΔH &= ΣΔH_{f^{o}merchandise} − Σ ΔHf_{f^{o}reactants} &= -285.8 + 291.83 &= 6.03 kJ mol^{-1} finish{align*} |
(b)
| start{align*} textual content{At melting level}, 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 Okay^{-1} mol^{-1} finish{align*} |
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