JK-TK/Newdata huggingface dataset:
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. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure.
∆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. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory.
The ammonium nitrate(V)would take less time to dissolves. Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation.
Increase in temperature reduces lattice energy causing endothermic dissolution to be faster (g)Illustrate the process above in an energy level diagram NH4+ (g) + NO3-(g) +∆H NH4+ (aq)+NO3-(aq) Energy(kJ) +∆H ∆H = -22.176kJ NH4NO3-(s) Reaction path /progress/coordinate (h) 100cm3 of distilled water at 25oC was added carefully 3cm3 concentrated sulphuric(VI)acid of density 1.84gcm-3.The temperature of the mixture rose from 250C to 38oC.Calculate the molar heat of solution of sulphuric(VI)acid (S=32.0,H=1.0,0=16.0) Working
30 Molar mass of H2SO4 = 98g Mass of H2SO4= Density x volume => 1.84gcm-3 x 3cm3 = 5.52 g Mass of H2O = Density x volume => 1.00gcm-3 x 100cm3 = 100 g Moles of H2SO4= mass => 5.52 g = 0.0563 moles Molar mass of H2SO4 98g Enthalpy change ∆H= (mass of acid + water) x specific heat capacity of water x ∆T => (100 +5.52 g) x 4.2 x 13oC = 5761.392 J = 5.761392 kJ 1000 ∆Hs of H2SO4= ∆H => 5.761392 kJ = -102.33378kJmoles-1 Moles of H2SO4 0.0563 moles (e)Standard enthalpy/heat of formation ∆Hᶿf The molar enthalpy of formation ∆Hᶿf is defined as the energy change when one mole of a compound is formed from its elements at 298K(25oC) and 101325Pa(one atmosphere)pressure. ∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”.
∆Hᶿf is practically difficult to determine in a school laboratory. It is determined normally determined by applying Hess’ law of constant heat summation. Hess’ law of constant heat summation states that “the total enthalpy/heat/energy change of a reaction is the same regardless of the route taken from reactants to products at the same temperature and pressure”. Hess’ law of constant heat summation is as a result of a series of experiments done by the German Scientist Henri Hess(1802-1850). He found that the total energy change from the reactants to products was the same irrespective of the intermediate products between. i.e. A(s) --∆H1-->C(s) = A(s) --∆H2-->B(s)--∆H3-->C(s) Applying Hess’ law of constant heat summation then:
31 A(s) ∆H2 B(s) ∆H1 ∆H3 C(s) The above is called an energy cycle diagram. It can be used to calculate any of the missing energy changes since: (i) ∆H1 =∆H2 + ∆H3 (ii) ∆H2 =∆H1 + -∆H3 (iii) ∆H3 = - ∆H1 + ∆H2 Examples of applying Hess’ law of constant heat summation 1.Calculate the molar enthalpy of formation of methane (CH4) given that ∆Hᶿc of carbon-graphite is -393.5kJmole-1,Hydrogen is -285.7 kJmole-1 and that of methane is 890 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form methane. Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.
Methane will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. C(s)+ 2H2 (g)+2O2 (g) --∆H1--> CH4(g) +2O2(g) --∆H2--> CO2(g)+2H2O(l) C(s)+ 2H2 (g)+2O2 (g) --∆H3--> CO2(g)+2H2O(l) Energy cycle diagram C(s) + 2H2 (g) + 2O2(g) ∆H1=∆Hᶿc =-890.4kJ CH4(g)+2O2(g) ∆H3=∆Hᶿc =-393.5kJ ∆H3=∆Hᶿc =-285.7kJ x 2 ∆H2= ∆Hᶿf= x CO2(g) + 2H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 -393.5 + (-285.7 x 2) = -890.4kJ + x x = -74.5 kJ
32 Heat of formation ∆Hᶿf CH4 = -74.5 kJmole-1 2. Calculate the molar enthalpy of formation of ethyne (C2H2) given : ∆Hᶿc of carbon-graphite = -394kJmole-1,Hydrogen = -286 kJmole-1 , (C2H2) = -1300 kJmole-1 Working Carbon-graphite ,hydrogen and oxygen can react to first form ethyne. Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water.
Ethyne will then burn in the oxygen present to form carbon(IV)oxide and water. Carbon-graphite can burn in the oxygen to form carbon(IV)oxide. Hydrogen can burn in the oxygen to form water. 2C(s)+ H2 (g)+2 ½ O2 (g) --∆H1--> C2 H2 (g) +2 ½ O2(g) --∆H2--> CO2(g)+H2O(l) 2C(s)+ H2 (g)+ 2 ½ O2 (g) --∆H3--> 2CO2(g)+H2O(l) Energy cycle diagram 2C(s) + H2 (g) +2½O2(g) ∆H1=∆Hᶿf =x C2 H2+2½O2(g) ∆H3=∆Hᶿc =-394kJx 2 ∆H3=∆Hᶿc =-286kJ ∆H2= ∆Hᶿc= -1300kJ 2CO2(g) + H2O(l) Substituting: ∆H3 = ∆H1 + ∆H2 ( -394 x 2) + -286 = -1300kJ + x x = +244 kJ Heat of formation ∆Hᶿf CH4 = +244 kJmole-1 3. Calculate the molar enthalpy of formation of carbon(II)oxide (CO) given : ∆Hᶿc of carbon-graphite = -393.5kJmole-1, ∆Hᶿc of carbon(II)oxide (CO)= -283 kJmole-1 Working Carbon-graphite reacts with oxygen first to form carbon (II)oxide (CO). Carbon(II)oxide (CO) then burn in the excess oxygen to form carbon(IV)oxide. Carbon-graphite can burn in excess oxygen to form carbon (IV) oxide.
Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table 1. Table 1:Sample results
35 Experiment I II Highest /lowest temperature T2 27.0 29.0 Initial temperature T1 24.0 25.0 Change in temperature ∆T 3.0 4.0 Experiment II Weigh accurately 8.0g of anhydrous copper(II)sulphate(VI). Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II.
Measure 100cm3 of distilled water into a beaker. Determine its temperature T1 .Put all the crystals of the copper(II)sulphate(VI)pentahydrate carefully into the beaker. Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II.
Stir using a thermometer and determine the highest temperature change T2 Repeat the procedure again to complete table II. Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4.
Table II :Sample results Experiment I II Highest /lowest temperature T2 26.0 27.0 Initial temperature T1 25.0 25.0 Change in temperature ∆T 1.0 2.0 Questions (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => 3.0 +4.0 = 3.5 oC 2 (ii)Table II ∆T= T2 -T1 => 1.0 +2.0 = 1.5 oC 2 (b)Calculate the number of moles of solid used in: (i)Experiment I Moles of CuSO4.5H2O = Mass => 12.5 = 0.05 moles Molar mass 250 (ii)Experiment II Moles of CuSO4 = Mass => 8.0 = 0.05 moles Molar mass 160 (c)Calculate the enthalpy change for the reaction in: (i)Experiment I Enthalpy change of CuSO4.5H2O= mass of Water(m) x c x ∆T =>100cm3 x 4.2 x 3.5 oC = -1.47kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T
36 =>100cm3 x 4.2 x 1.5 oC = -0.63kJ 1000 (c)Calculate the molar enthalpy of solution CuSO4 .5H2O (s) form the results in (i)experiment I. ∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information.
∆Hs = CuSO4.5H2O= ∆H => -1.47kJ = 29.4kJ Number of Moles 0.05 moles (ii)experiment II. ∆Hs = CuSO4= ∆H => -0.63kJ = 12.6kJ Number of Moles 0.05 moles (d) Using an energy level diagram, calculate the molar enthalpy change for the reaction: CuSO4 .5H2O (s) -> CuSO4(s) + 5H2O(l) Energy cycle diagram CuSO4(s) + (aq) + 5H2O(l) ∆H1=x CuSO4. 5H2O (s)+ (aq) ∆H3= =-29.4kJ ∆H2= -12.6kJ CuSO4 .5H2O (aq) ∆H3 = ∆H1 +∆H2 =>-29.4kJ = -12.6kJ + x =>-29.4kJ - (+12.6kJ) = x x = 16.8kJ b)Practical example II Determination of enthalpy of solution of ammonium chloride Theoretical information. Ammonium chloride dissolves in water to form ammonium chloride solution. Aqueous ammonia can react with excess dilute hydrochloric acid to form ammonium chloride solution. The heat change taking place can be calculated from the heat of reactions:
37 (i) NH3(aq) + HCl(aq) -> NH4Cl(s) (ii) NH4Cl(s) + (aq) -> NH4Cl(aq) (iii) NH3(aq) + HCl(aq) -> NH4Cl(aq) Experiment procedure I Measure 50cm3 of water into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table I. Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I.
Add exactly 5.0g of ammonium chloride crystals weighed carefully into the water. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table I. Sample results TableI Experiment I II final temperature(oC) 19.0 20.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 3.0 2.0 Experiment procedure II Measure 25cm3 of 2M aqueous ammonia into a 100cm3 beaker. Record its temperature T1 as initial temperature to the nearest 0.5oC in table II. Measure 25cm3 of 2M hydrochloric acid solution. Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II.
Add the acid into the beaker containing aqueous ammonia. Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ.
Stir and record the highest temperature change T2 as the final temperature change. Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:
39 1.
Repeat the above procedure to complete table II. Sample results:Table II Experiment I II final temperature(oC) 29.0 29.0 initial temperature(oC) 22.0 22.0 temperature change ∆T(oC) 7.0 7.0 Sample Calculations: (a)Calculate the average ∆T in (i)Table I ∆T= T2 -T1 => -3.0 +-2.0 = 2.5 oC 2 (ii)Table II ∆T= T2 -T1 => 7.0 +7.0 = 7.0 oC 2 (b)Calculate the enthalpy change for the reaction in:
38 (i)Experiment I Enthalpy change ∆H = mass of Water(m) x c x ∆T =>50cm3 x 4.2 x 2.5 oC = +0.525kJ 1000 (ii)Experiment II Enthalpy change of CuSO4 = mass of water(m) x c x ∆T =>25+25cm3 x 4.2 x 7 oC = +1.47kJ 1000 (c)Write the equation for the reaction taking place in: (i)Experiment I NH4Cl(s) + (aq) -> NH4Cl(aq) (ii)Experiment I NH3(aq) + HCl(aq) -> NH4Cl(aq) (d)Calculate the enthalpy change ∆H for the reaction: NH3(g) + HCl(g) -> NH4Cl(s) given that: (i) NH3(g) + (aq) -> NH3(aq) ∆H= -40.3kJ (ii) (aq) + HCl(g) -> HCl(aq) ∆H= -16.45kJ (e)Applying Hess’ Law of constant heat summation: Energy level diagram N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf NH4Cl(s) + aq +0.525kJ=∆H4 (aq) (aq) - 40.3kJ=∆H1 -16.43kJ=∆H2 NH3 (aq) + HCl(aq) -1.47kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 40.3kJ + -16.43kJ + -1.47kJ = +0.525kJ + ∆Hf =>∆Hf = -58.865kJ. Practice theoretical examples:
39 1. Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that.
Using an energy level diagram calculate the ∆Hs of ammonium chloride crystals given that. ∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
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20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed.
∆Hf of NH3 (aq) = -80.54kJ mole-1 ∆Hf of HCl (aq) = -164.46kJ mole-1 ∆Hf of NH4Cl (aq) = -261.7483kJ mole-1 ∆Hs of NH4Cl (aq) = -16.8517kJ mole-1 N2(g) + 1½ H2(g) + ½ Cl2 ∆Hf=-261.7483kJ NH4Cl(s) + aq x=∆Hs (aq) (aq) - 80.54kJ=∆H1 -164.46kJ=∆H2 NH3 (aq) + HCl(aq) 16.8517kJ=∆H3 NH4Cl(s) ∆H1 + ∆H2 + ∆H3 = ∆H4 + ∆Hf - 80.54kJ + -164.46kJ + -16.8517kJ = -261.7483kJ + ∆Hf =>∆Hf = -33.6kJmole-1. Study the energy cycle diagram below and use it to: (a)Identify the energy changes ∆H1 ∆H2 ∆H3 ∆H4 ∆H5 ∆H6
40 ∆H1 - enthalpy/heat of formation of sodium chloride (∆Hf) ∆H2 -enthalpy/heat of atomization of sodium (∆Hat) ∆H3 -enthalpy/heat of ionization/ionization energy of sodium (∆H i) ∆H4 -enthalpy/heat of atomization of chlorine (∆Hat) ∆H5 -enthalpy/heat of electron affinity of chlorine (∆He) ∆H6 enthalpy/heat of lattice/Lattice energy of sodium chloride(∆H l) (b) Calculate ∆H1 given that ∆H2 =+108kJ , ∆H3=+500kJ, ∆H4 =+121kJ ,∆H5 =-364kJ and ∆H6 =-766kJ Working: ∆H1 =∆H2 +∆H3 +∆H4 +∆H5 +∆H6 Substituting: ∆H1= +108kJ + +500kJ + +121kJ +-364kJ + -766kJ ∆H1= -401kJmole-1 (c) Given the that: (i) Ionization energy of sodium = + 500kJmole-1 (ii)∆Hat of sodium = + 110kJmole-1 (iii) Electron affinity of chlorine = - 363kJmole-1 (iv)∆Hat of chlorine = + 120kJmole-1 (v) ∆Hf of sodium chloride = -411kJ , calculate the lattice energy of sodium chloride using an energy cycle diagram. 41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
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20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up.