Energetics Topic 5.1 5.2
Topic 5.1 Exothermic and Endothermic Reactions??
total energy of the universe is a constant if a system loses energy, it must be gained by the surroundings, and vice versa
Enthalpy H a measure of the heat content (not temp) you cannot measure the actual enthalpy of a substance but you can measure an enthalpy CHANGE because of energy it takes in or releases = Greek letter delta meaning change H = heat. so, H means heat change.
Why a standard? enthalpy values vary according to the conditions a substance under these conditions is said to be in its standard state pressure: 100 kpa (1 atmosphere) temperature: usually 298K (25 C) if these were not standardized, then energy could be leaving or entering the system modify the symbol from Enthalpy Change Standard Enthalpy Change (at 298K)
Enthalpy (Heat) of Reaction H = H products H reactants lower energy is more stable
Exothermic reactions heat energy is given out by the reaction hence the surroundings increase in temperature (feels hot) occurs when bonds are formed new products are more stable and extra energy is given off H products < H reactants H is negative examples combustion of fuels respiration neutralization reactions (acid reacts with something)
ENTHALPY activation energy energy necessary to get the reaction going reactants energy given out, H is negative products REACTION CO-ORDINATE
H 2 + Cl 2 2HCl Energy taken in to break bonds. H, H, Cl, Cl (Atoms) Energy given out when bonds are made. energy H-H, Cl-Cl Reactants Overall energy change, H H-Cl, H-Cl Products
H 2 + Cl 2 2HCl Energy in = +678kJ H, H, Cl, Cl (Atoms) Energy out = -862kJ energy H-H, Cl-Cl Reactants Overall energy change, H = -184kJ H-Cl, H-Cl Products
Endothermic reactions heat energy is taken in by the reaction mixture hence the surroundings decrease in temperature (feels cold) occurs when bonds are broken the reactants were more stable (bonds are stronger) therefore, took energy from the surrounding to break bonds H reactants < H product H is positive examples
ENTHALPY activation energy energy necessary to get the reaction going products energy taken in, H is positive reactants REACTION CO-ORDINATE
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Summary Table Exothermic reactions Energy is given out to the surroundings H is negative Products have less energy than reactants Endothermic reactions Energy is taken in from the surroundings H is positive Products have more energy than reactants
Topic 5.2 Calculation of enthalpy change Notice less sources of error here compared to our lab
calorimetry measurement of heat flow calorimeter device used to measure heat flow heat energy that is transferred from one object to another due to a difference in temperature measures total energy in a given substance temperature a measure of the average kinetic energy of a substance regardless how much is there
Temperature vs. Heat 50 ml water 100 ml water 100 C 100C 100ml of water contains twice the heat of 50 ml.
19 Heat Capacity/Specific Heat the amount of energy a substance absorbs depends on: mass of material temperature kind of material and its ability to absorb or retain heat. heat capacity the amount of heat required to raise the temperature of a substance 1 o C (or 1 Kelvin) molar heat capacity the amount of heat required to raise the temperature of one mole 1 o C (or 1 Kelvin) specific heat the amount of heat required to raise the temperature of 1 gram of a substance 1 o C (or 1 Kelvin)
20 Specific Heat (c) values for Some Common Substances Substance J g -1 K -1 Water (liquid) 4.184 Water (steam) 2.080 Water (ice) 2.050 Copper 0.385 or kj kg -1 K -1 if multiply by 1000 Aluminum 0.897 Ethanol 2.44 Lead 0.127
Heat energy change q = m c T q = change in heat (same as H if pressure held constant) m = mass in grams or kilograms c = specific heat in J g -1 K -1 or kj kg -1 K -1 (or Celsius which has same increments as Kelvin) T = temperature change (final initial)
Measuring the temperature change in a calorimetry experiment can be difficult since the system is losing heat to the surroundings even as it is generating heat. By plotting a graph of time vs. temperature it is possible to extrapolate back to what the maximum temperature would have been had the system not been losing heat to the surroundings.
Heat Transfer Problem 1 Calculate the heat that would be required to heat an aluminum cooking pan whose mass is 402.5 grams, from 20.5 o C to 201.5 o C. The specific heat of aluminum is 0.902 J g -1 o C -1. q = mc T only 4 sig. figs. = (402.5 g) (0.902 J g -1 o C -1 )(181.0 o C) = 65,712.955 J = 65,710 J with correct sig. figs.
Heat Transfer Problem 2 What is the final temperature when 50.15 grams of water at 20.5 o C is added to 80.65 grams water at 60.5 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is 4.184 J g -1 o C -1 Solution: q (Cold) = q ( hot ) so mc T = mc T Let T = final temperature (50.15 g) (4.184 J g -1 o C -1 )(T- 20.5 o C) = (80.65 g) (4.184 J g -1 o C -1 )(60.5 o C- T) (50.15 g)(t- 20.5 o C) = (80.65 g)(60.5 o C- T) 50.15T -1030 = 4880 80.65T 130.80T = 5830 T = 44.6 o C
Heat Transfer Problem 3 On complete combustion, 0.18g of hexane raised the temperature of 100.5g water from 22.5 C to 47.5 C. Calculate its enthalpy of combustion. Heat absorbed by the water q = mc T q = 100.5 (4.18) (25.0) = 10,500 J which is same as 10.5 kj Moles of hexane burned = mass / molar mass = 0.18 g / 86 g/mol = 0.0021 moles of hexane Enthalpy change means find heat energy / mole = 10.5 kj/ 0.0021 mol = 5000 kj mol -1 or 5.0 x 10 3 kj mol -1 hexane is C 6 H 14