chemistry m4 energetics
TRANSCRIPT
OBJECTIVES
At the end of the lesson you should be able to:1. describe energy changes in bond formation and bond breaking2. distinguish between exothermic and endothermic reactions in terms of
energy 3. explain energy diagrams representing endothermic and exothermic
reactions4. explain the following terms: activation energy, enthalpy change 5. define heat of solution 6. calculate the heat of solution from experiments or from experimental
data 7. define heat of neutralisation 8. calculate the heat of neutralisation from experiments or from
experimental data
ENDOTHERMIC REACTION
If the energy (heat) absorbed is greater than the energy (heat) released, then the overall reaction is ENDOTHERMIC.
Energy Profile diagram for an endothermic reaction
EXOTHERMIC REACTION
If the energy (heat) released is greater than the energy (heat) absorbed, then the overall reaction is EXOTHERMIC.
Energy Profile diagram for an exothermic reaction
COMPARING ENERGY PROFILE DIAGRAMS
Compare both energy profile diagrams. State the differences you observe
Energy Profile diagram for: An endothermic reaction An exothermic
reaction
DIFFERENCES BETWEEN ENERGY PROFILES
ENDOTHERMIC REACTION EXOTHERMIC REACTION
Heat content of the products is greater than that of the reactants
Heat content of the products is less than that of the reactants
Heat (enthalpy) change, ∆H is positive (+∆H)
Heat (enthalpy) change, ∆H is negative (-∆H)
Measuring the heat (enthalpy) change, ∆H
∆H = m c ∆Twhere: • m is the mass of the solution• c is the specific heat capacity• ∆T is the change in temperature
HEAT OF SOLUTION
• The heat absorbed or evolved when one mole of a substance is dissolved in an infinite amount of solvent so that further dilution causes no further heat change.
• Go to the link below and note the instructions before closing as you will do this as a planning and designing lab.http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/heat_soln.html
• We will now use this link to calculate heat of solution.
HEAT OF SOLUTION: PROBLEM
1. Add 3.00 g of NaOH to 100 cm3 of water.2. Note the initial temperature and the highest
or lowest temperature reached. (If you are not sure, the change in temperature is given on the graph.)
3. Calculate the heat of solution of NaOH. (specific heat capacity of water is 4.2 J g-1 K-1 and the density of water is 1 g cm-3)
ACTIVITY
Try these using the same programme!
• 2.5g NH4Cl in 20 cm3 of water.
• 4.0 g Na2CO3 in 40 cm3 of water
HEAT OF NEUTRALISATION: PROBLEM
The heat absorbed or evolved when one mole of water is produced when an acid reacts with a base.
Problem:• When 50 cm3 of 2 mol dm-3 hydrochloric acid at 25 oC is
reacted with 50 cm3 of 2 mol dm-3 sodium hydroxide at 25 oC, the temperature increased to 36 oC. Determine the heat of neutralisation.
• Assume that the density and the specific heat capacity of the solution is the same as that for water.
Answer to Problem
Initial Temperature = 25 oC (average of both solutions)
Final Temperature = 36 oCDensity of solution = mass/volumeMass = volume x density
= 100 cm3 x 1 g cm-3
= 100 g
Answer to Problem (continued)
Calculation of the heat released:
∆H = m c ∆T
∆H = 100 x 4.2 x (36 – 25)
= 4620 J
Answer to Problem (continued)
Calculation of the number of moles of reactants:Number of moles of HCl (and NaOH):
1000 cm3 contain 2 moles
50 cm3 contain x
x = 50 x 2
1000
= 0.1 moles
NaOH + HCl → NaCl + H2O
1 mol NaOH ≡ 1 mol HCl → 1 mol H2O
0.1 mol NaOH ≡ 0.1 mol HCl → 0.1 mol H2O