Topic 5.1 Measuring Energy Changes

total energy of the universe is a constant law of conservation of energy – if a system loses energy, it must be gained by the surroundings, and vice versa

when examining energy changes in a chemical reactions, we divide “the universe” into two parts

Temperature vs. Heat temperature – a measure of the average kinetic energy of the particles regardless of the amount heat (q) – one of many forms of energy – the transfer from a warmer body to a cooler body as a result of a temperature gradient – measures the total energy in a given substance (amount does matter)

50 ml water 100 C 100 ml water 100C 100ml of water contains twice the heat of 50 ml.

the amount of heat used or released (change) in a system a state function- meaning a quantity that only depends on the state at the moment –i.e. temp, pressure, volume –the value is independent of changes “along the way” –only the beginning and end matter you cannot measure the actual enthalpy of a substance directly –but you can measure an enthalpy CHANGE because of energy it takes in or releases H Enthalpy

use the symbol a substance under these conditions is said to be in its standard state –pressure: 100 kPa (1 atmosphere) –temperature: 298K (25°C) vs

Enthalpy (Heat) of a Reaction the amount of energy absorbed or given off in a chemical reaction  H = ∑  H products − ∑  H reactants

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 –  is negative examples –combustion of fuels –respiration –neutralization reactions (acid reacts with something) Exothermic reactions

REACTION CO-ORDINATE ENTHALPY energy given out, ∆H is negative reactants products activation energy energy necessary to get the reaction going

H 2 + Cl 2  2HCl energy H-H, Cl-Cl Reactants H, H, Cl, Cl (Atoms) H-Cl, H-Cl Products Energy taken in to break bonds. Energy given out when bonds are made. Overall energy change, H

H 2 + Cl 2  2HCl energy H-H, Cl-Cl Reactants H, H, Cl, Cl (Atoms) H-Cl, H-Cl Products Energy out = -862kJ Overall energy change, H = -184kJ Energy in = +678kJ

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) overall, took energy from the surroundings H reactants < H product  is positive examples – photosynthesis Endothermic reactions

REACTION CO-ORDINATE ENTHALPY reactants products Energy taken in and now stored in the bonds. ∆H is positive activation energy energy necessary to get the reaction going

Summary Table Exothermic reactions Endothermic reactions Energy is given out to the surroundings Energy is taken in from the surroundings ∆H is negative∆H is positive Products have less energy than reactants Products have more energy than reactants

Calculation of enthalpy change

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)

Substance J g -1 K -1 Water (liquid)4.184 Water (steam)2.080 Water (ice)2.050 Copper0.385 Aluminum0.897 Ethanol2.44 Lead0.127 Specific Heat (c) values for Some Common Substances 21 or kJ kg -1 K -1 if multiply by 1000

Heat energy change/transfer q = m c  T q = change in heat (same as  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

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 grams, from 20.5 o C to o C. The specific heat of aluminum is J g -1 o C -1. q = mc  T = ( g) (0.902 J g -1 o C -1 )(181.0 o C ) = 65, J = 65,710 J with correct sig. figs. only 4 sig. figs.

Heat Transfer Problem 2 What is the final temperature when grams of water at 20.5 o C is added to grams water at 60.5 o C? Assume that the loss of heat to the surroundings is negligible. The specific heat of water is J g -1 o C -1 Solution:  q (cold) =  q (hot) so… mC  T = mC  T Let T f = final temperature of the water (50.15 g) (4.184 J g -1 o C -1 )(T f o C) = (80.65 g) (4.184 J g -1 o C -1 )(60.5 o C - T f ) (50.15 g)(T f o C) = (80.65 g)(60.5 o C- T f ) 50.15T f – 1030 = 4880 – 80.65T f T f = 5910 T f = 45.2 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 in kJ mole -1 of hexane. Heat absorbed by the water… q = mc  T q = (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 = moles of hexane –Need to find heat energy / mole = 10.5 kJ/ mol = 5000 kJ mol -1 or 5.0 x 10 3 kJ mol -1 hexane is C 6 H 14