JK-TK/Newdata huggingface dataset:
Carefully add all the alkali into the calorimeter/200cm3 plastic beaker wrapped in cotton wool/tissue paper containing the acid. Stir vigorously the mixture with the thermometer. Determine the highest temperature change to the nearest 0.5oC T4 as the final temperature of the mixture. Repeat the experiment to complete table 1. Table I . Sample results Experiment I II Temperature of acid T1 (oC) 22.5 22.5 Temperature of base T2 (oC) 22.0 23.0 Final temperature of solution T4(oC) 35.5 36.0 Initial temperature of solution T3(oC) 22.25 22.75
22 (a)Calculate T6 the average temperature change T6 = 13.25 +13.75 = 13.5 oC 2 (b)Why should the apparatus be very clean? Impurities present in the apparatus reacts with acid /base lowering the overall temperature change and hence ∆Hᶿn. (c)Calculate the: (i)number of moles of the acid used number of moles = molarity x volume => 2 x 50 = 0.1moles 1000 1000 (ii)enthalpy change ∆H of neutralization. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above.
∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T(T6) => (50 +50) x 4.2 x 13.5 = 5670Joules = 5.67kJ (iii) the molar heat of neutralization the acid. ∆Hn = Enthalpy change ∆H => 5.67kJ = 56.7kJ mole-1 Number of moles 0.1moles (c)Write the ionic equation for the reaction that takes place OH-(aq)+ H+(aq) -> H2O(l) (d)The theoretical enthalpy change is 57.4kJ. Explain the difference with the results above. The theoretical value is higher Heat/energy loss to the surrounding/environment lowers ∆T/T6 and thus ∆Hn Heat/energy is absorbed by the reaction vessel/calorimeter/plastic cup lowers ∆T and hence ∆Hn (e)Compare the ∆Hn of the experiment above with similar experiment repeated with neutralization of a solution of: (i) potassium hydroxide with nitric(V) acid The results would be the same/similar. Both are neutralization reactions of strong acids and bases/alkalis that are fully /wholly dissociated into many free H+ / H3O+ and OH- ions. (ii) ammonia with ethanoic acid Temperature change( T5) 13.25 13.75
23 The results would be lower/∆Hn would be less. Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress Theoretical examples 1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide.
Both are neutralization reactions of weak acids and bases/alkalis that are partially /partly dissociated into few free H+ / H3O+ and OH- ions. Some energy is used to ionize the molecule. (f)Draw an energy level diagram to illustrate the energy changes H2 H+ (aq)+OH (aq) Energy (kJ) ∆H = -56.7kJ H1 H2O (l) Reaction path/coordinate/progress Theoretical examples 1.The molar enthalpy of neutralization was experimentary shown to be 51.5kJ per mole of 0.5M hydrochloric acid and 0.5M sodium hydroxide. If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC.
If the volume of sodium hydroxide was 20cm3, what was the volume of hydrochloric acid used if the reaction produced a 5.0oC rise in temperature? Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube.
Working: Moles of sodium hydroxide = molarity x volume => 0.5 M x 20cm3 = 0.01 moles 1000 1000 Enthalpy change ∆H = ∆Hn => 51.5 = 0.515kJ Moles sodium hydroxide 0.01 moles Mass of base + acid = Enthalpy change ∆H in Joules Specific heat capacity x ∆T => 0.515kJ x 1000 = 24.5238g 4.2 x 5 Mass/volume of HCl = Total volume – volume of NaOH =>24.5238 - 20.0 = 4.5238 cm3 3. ∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker.
∆Hn of potassium hydroxide was practically determined to be 56.7kJmole-1.Calculate the molarity of 50.0 cm3 potassium hydroxide used to neutralize 25.0cm3 of dilute sulphuric(VI) acid raising the temperature of the solution from 10.0oC to 16.5oC. ∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker. Record its initial temperature at volume of base =0.
∆H = (m)mass of solution(acid+base) x (c)specific heat capacity of solution x ∆T
24 => (50 +25) x 4.2 x 6.5 = 2047.5Joules Moles potassium hydroxide =Enthalpy change ∆H ∆Hn 2047.5Joules = 0.0361 moles 56700Joules Molarity of KOH = moles x 1000 => 0.0361 moles x 1000 = 0.722M Volume used 50cm3 3.Determine the specific heat capacity of a solution of a solution mixture of 50.0cm3 of 2M potassium hydroxide neutralizing 50.0cm3 of 2M nitric(V) acid if a 13.25oC rise in temperature is recorded.(1mole of potassium hydroxide produce 55.4kJ of energy) Moles of potassium hydroxide = molarity KOH x volume 1000 => 2 M x 50cm3 = 0.1 moles 1000 Enthalpy change ∆H = ∆Hn x Moles potassium hydroxide => 55.4kJ x 0.1 moles = 5.54kJ x 1000=5540Joules Specific heat capacity = Enthalpy change ∆H in Joules Mass of base + acid x ∆T => 5540 = 4.1811J-1g-1K-1 (50+50) x 13.25 Graphically ∆Hn can be determined as in the example below: Procedure Place 8 test tubes in a test tube rack .Put 5cm3 of 2M sodium hydroxide solution into each test tube. Measure 25cm3 of 1M hydrochloric acid into 100cm3 plastic beaker. Record its initial temperature at volume of base =0. Put one portion of the base into the beaker containing the acid. Stir carefully with the thermometer and record the highest temperature change to the nearest 0.5oC. Repeat the procedure above with other portions of the base to complete table 1 below Table 1:Sample results.
(b)Plot a graph of volume of sodium hydroxide against temperature change. 28.7=T2 temperature(oC) ∆T From the graph show and determine : (i)the highest temperature change ∆T ∆T =T2-T1 => highest temperature-T2 (from extrapolating a correctly plotted graph) less lowest temperature at volume of base=0 :T1 =>∆T = 6.7 – 0.0 = 6.70C (ii)the volume of sodium hydroxide used for complete neutralization From a correctly plotted graph – 16.75cm3 (c)Calculate the number of moles of the alkali used Moles NaOH = molarity x volume =>2M x 16.75cm3 = 0.0335 moles 1000 1000 (d)Calculate ∆H for the reaction ∆H = mass of solution(acid+base) x c x ∆T =>(25.0 + 16.75) x 4.2 x 6.7 = 1174.845 J = 1.174845kJ 1000 (e)Calculate the molar enthalpy of neutralization of the alkali. ∆Hn = ∆Hn = 1.174845kJ = 35.0701kJ Number of moles 0.0335 Volume of sodium hydroxide(cm3) 22. oC T1
26 (d)Standard enthalpy/heat of solution ∆Hᶿs The standard enthalpy of solution ∆Hᶿsis defined as the energy change when one mole of a substance is dissolve in excess distilled water to form an infinite dilute solution. An infinite dilute solution is one which is too dilute to be diluted further. Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl).
Dissolving a solid involves two processes: (i) breaking the crystal of the solid into free ions(cations and anion).This process is the opposite of the formation of the crystal itself. The energy required to form one mole of a crystal structure from its gaseous ions is called Lattice energy/heat/enthalpy of lattice (∆Hl). Lattice energy /heat/enthalpy of lattice (∆Hl) is an endothermic process (+∆Hl). The table below shows some ∆Hl in kJ for the process MX(s) -> M+ (g) + X- (g) Li Na K Ca Mg F +1022 +900 +800 +760 +631 Cl +846 +771 +690 +2237 +2493 Br +800 +733 +670 +2173 +2226 (ii)surrounding the free ions by polar water molecules. This process is called hydration. The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water.
The energy produced when one mole of ions are completely hydrated is called hydration energy/ heat/enthalpy of hydration(∆Hh).Hydration energy /enthalpy of hydration(∆Hh) is an exothermic process(∆Hh). The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature.
The table below shows some ∆Hh in kJ for some ions; ion Li+ Na+ K+ Mg2+ Ca2+ F- Cl- Br- ∆Hh -1091 -406 -322 -1920 -1650 -506 -364 -335 The sum of the lattice energy +∆Hl (endothermic) and hydration energy -∆Hh (exothermic) gives the heat of solution-∆Hs ∆Hs = ∆Hl +∆Hh Note Since ∆Hl is an endothermic process and ∆Hh is an exothermic process then ∆Hs is: (i)exothermic if ∆Hl is less than ∆Hh and hence a solid dissolve easily in water. (ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic .
(ii)endothermic if ∆Hl is more than ∆Hh and hence a solid does not dissolve easily in water. (a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature.
(a)Dissolving sodium chloride crystal/s: (i) NaCl ----breaking the crystal into free ions---> Na +(g)+ Cl-(g) ∆Hl =+771 kJ
27 (ii) Hydrating the ions; Na +(g) + aq -> Na(aq) ∆Hh = - 406 kJ Cl-(g) + aq -> Cl-(aq) ∆Hh = - 364 kJ ∆Hs =∆Hh +∆Hs -> (- 406 kJ + - 364 kJ) + +771 kJ = + 1.0 kJmole-1 NaCl does not dissolve easily in water because overall ∆Hs is endothermic Solubility of NaCl therefore increases with increase in temperature. Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic .
Increase in temperature increases the energy to break the crystal lattice of NaCl to free Na +(g)+ Cl-(g) (b)Dissolving magnesium chloride crystal/s// MgCl2 (s) ->MgCl2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Cl-(g) ∆Hl =+2493 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Cl-(g) + aq -> 2Cl-(aq) ∆Hh = (- 364 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 364 x 2 kJ)) + +2493 kJ = -155.0 kJmole-1 MgCl2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature.
Solubility of MgCl2 (s) therefore decreases with increase in temperature. (c)Dissolving Calcium floride crystal/s// CaF2 (s) -> CaF2 (aq) (i) CaF2 -->Ca 2+(g)+ 2F-(g) ∆Hl =+760 kJ (ii) Hydrating the ions; Ca 2+(g) + aq -> Ca 2+(g) (aq) ∆Hh = - 1650 kJ 2F-(g) + aq -> 2F-(aq) ∆Hh = (- 506 x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1650 kJ + (- 506 x 2 kJ)) + +760 kJ = -1902.0 kJmole-1 CaF2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of CaF2 (s) therefore decreases with increase in temperature. (d)Dissolving magnesium bromide crystal/s// MgBr2 (s) ->MgBr2 (aq) (i) MgCl2 --breaking the crystal into free ions-->Mg 2+(g)+ 2Br-(g) ∆Hl =+2226 kJ (ii) Hydrating the ions; Mg 2+(g) + aq -> Mg 2+(g) (aq) ∆Hh = - 1920 kJ 2Br-(g) + aq -> 2Br-(aq) ∆Hh = (- 335x 2) kJ ∆Hs =∆Hh +∆Hs -> (- 1920 kJ + (- 335 x 2 kJ)) + +2226 kJ = -364.0 kJmole-1 MgBr2 (s) dissolve easily in water because overall ∆Hs is exothermic . Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined.
Solubility of MgBr2(s) therefore decreases with increase in temperature. 28 Practically the heat of solution can be determined from dissolving known amount /mass/volume of solute in known mass /volume of water/solvent. From the temperature of solvent before and after dissolving the change in temperature(∆T) during dissolution is determined. To determine the ∆Hs ammonium nitrate Place 100cm3 of distilled water into a plastic beaker/calorimeter. Determine its temperature and record it at time =0 in table I below. Put all the 5.0g of ammonium nitrate (potassium nitrate/ammonium chloride can also be used)provided into the plastic beaker/calorimeter, stir using a thermometer and record the highest temperature change to the nearest 0.5oCafter every ½ minute to complete table I. Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V).
Continue stirring the mixture throughout the experiment. Sample results: Table I Time (minutes) 0.0 ½ 1 1 ½ 2 2 ½ 3 3 ½ Temperature()oC 22.0 21.0 20.0 19.0 19.0 19.5 20.0 20.5 (a)Plot a graph of temperature against time(x-axis) 22.0=T1 temperature(oC) ∆T (b)From the graph show and determine the highest temperature change ∆T ∆T =T2-T1 => lowest temperature-T2 (from extrapolating a correctly plotted graph) less highest temperature at volume of base=0 :T1 Time (minutes) 18.7. oC T1
29 =>∆T =18.7 – 22.0 = 3.30C (c)Calculate the number of moles of ammonium nitrate(V) used Moles NH4NO3 = mass used => 5.0 = 0.0625 moles Molar mass 80 (d)Calculate ∆H for the reaction ∆H = mass of water x c x ∆T ->100 x 4.2 x 3.3 = +1386 J = +1.386kJ 1000 (e)Calculate the molar enthalpy of dissolution of ammonium nitrate(V). ∆Hs = ∆H = +1.386kJ = + 22.176kJ mole-1 Number of moles 0.0625 moles (f)What would happen if the distilled water was heated before the experiment was performed. The ammonium nitrate(V)would take less time to dissolves.