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# IB Chemistry Topic 15 energetics hl

## on Apr 18, 2014

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## IB Chemistry Topic 15 energetics hlDocument Transcript

• Topic 15: Energetics (HL) 15.1: Standard Enthalpy Changes of a Reaction 15.1.1: Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion. Standard State: The form an element/compound exists under standard conditions:  Pressure:101 kPA  Temperature: 25°C (or 298K) A. Standard Enthalpy change of Formation (∆Hf ϴ ): enthalpy change that occurs when one mole of the substance is formed from its elements in their standards states. B. Standard Enthalpy change of Combustion (∆Hc ϴ ): enthalpy change that occurs when one mole of the substance burns completely under standard conditions. 15.1.2: Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion. Example: combustion of methane CH4 + 2O2 → 2H2O + CO2 We want to find the (Energy in/kJ – Energy out/kJ) of the reaction to determine the Enthalpy change. We can calculate bond enthalpies using the data from the IB data booklet. Average bond enthalpies at 298K Energy in C-H = 413 kJ mol -1 However, there are four C-H bonds in methane, so the energy will be : 413kJmol -1 x 4 = 1652 kJmol -1 2O2 → O=O → 498kJmol -1 x 2 = 996kJmol -1 Total energy in = 996 + 1652 = 2648 kJmol -1 Energy out CO2 → 2x C=O → 746kJmol -1 x2 = 1492 kJmol -1 2H2O → 4 x H-O → 4 x464 = 1856kJmol -1 Total energy out = 3348 kJmol -1 Therefore: Enthalpy Change = Energy in – Energy out = -700kJmol -1 15.2: Born-Haber Cycle 15.2.1: Define and apply the terms lattice enthalpy and electron affinity Lattice Enthalpy(∆ ϴ lat): enthalpy change that occurs when one mole of a solid ionic compound is separated into gaseous ions under standard conditions.We measure lattice enthalpy using kJ mol -
• Electron Affinity(∆H ϴ e): enthalpy change when you add an electron to an atom in the gaseous state. Example: X (g) + e → X – (g) 15.2.2: Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds. Factors affecting the lattice enthalpy of different ionic compounds:  Size of the ions: If the ions are smaller, there will be a stronger attractive force between the protons of one ion and the electrons of another. Conversely, if the ions are bigger, the opposite will occur.  Size of the charge: The greater the charge, the greater the attractive forces, and the higher the lattice enthalpy. 15.2.3: Construct a Born–Haber cycle for group 1 and 2 oxides and chlorides, and use it to calculate an enthalpy change. We can construct a born haber cycle to calculate the enthalpy change  The Born Haber Cycle, just like Hess’s Law, is a method to calculate energy change that cannot be measured directly.  The Enthalpy Change of Formation, which is what we usually want to calculate, can easily be found by adding all the individual values of each step together.
• Enthalpy of Atomization (∆H ϴ atom):: Energy required to form one mole of gaseous atoms. Notice how the half mole of Fluorine became 1 mole in the Enthalpy of Atomization. Enthalpy of Ionization (∆H ϴ i): the amount of energy required to make 1 mole of gaseous metal ions.  Born Haber Cycle: The sum of the enthalpies on any two sides must equal the enthalpy change of the remaining side. Formula: ∆H ϴ f = ∆H ϴ atom + ∆H ϴ i + ∆H ϴ e - ∆H ϴ lat 15.2.4: Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character. As we discussed earlier, the Born Haber Cycle can be used to assess the Lattice Enthalpy and the Enthalpy of Formation of an ionic compound. When we do an experimental and extract results, and the results are very different from the theoretical values, then this may suggest that the compound has more covalent characteristics than ionic. Example: The theoretical value of NaCl is 766 KJ mol-1, wheres the experimental value of NaCl is 771 KJ mol-. There is a small difference between the two values so we can assume that there is a high degree of ionic characteristics in NaCl. 15.3: Entropy Entropy (S) is considered to be the amount of disorder in a system. Formula: ΔS = Sfinal – Sinitial 15.3.1: State and explain the factors that increase the entropy in a system  Changes of state: When the product changes from state from solid to liquid and lastly, to gas, the state of disorder increases.  Mixing and Dissolving: If you mix two substances together, you will be increasing the disorder in the system. Therefore, the disorder of aqueous solutions will be larger than that of a liquid.  Formation of gaseous products: In a reaction, if one of the products is a gas, then the disorder in the system will increase, as the disorder of a gas is typically larger than that of a solid or a liquid.  Increased movement of particles: If you heat a liquid or a gas, you are providing the system with more kinetic energy, hence increasing the disorder in the system.  Increased number of particles: This increases the disorder in the system.
• 15.3.2: Predict whether the entropy change (ΔS) for a given reaction or process is positive or negative. REMEMBER: Change ∆S solid liquid increase (+) solid gas increase (+) liquid gas increase (+) liquid solid decrease (-) gas  solid decrease (-) gas liquid decrease (-) Example of Entropy Predicting: 1) X (g) + Y (g) –> A (g) + B (g) 2) X (g) + Y (g) –> A (s) + A (l) 3) X (s) + Y( s) –> A (s) + A (g) 4) X (s) + Y( s) –> A (s) + A (s) Answers: 1) There is no change in state so the entropy change is zero, or very close to zero. 2) There is a change of state from gas to liquid/solid, so the change in entropy is negative, as the disorder in the system decreases. 3) Notice that although one of the substances is still a solid, substance B changed state to a gas, so the change in entropy is positive, as the disorder in the system increases 4) There is no change in state, so the change in entropy is zero. 15.3.3: Calculate the standard entropy change for a reaction (S°) using standard entropy values (S°).  Enthalpy is very difficult to measure directly.  However, the absolute entropy can be measured directly. We give the absolute entropy the letter (S). ΔS ѳ = ΣS ѳ products – ΣS ѳ reactants Let’s do an example 1.Calculate entropy change in the following reaction: 3H2(g) +N2 (g) –> 2NH3 (g) However, we need to know the values of some absolute entropies Hydrogen: 131 KJ -1 mol -1 Nitrogen: 192 JK -1 mol -1 Ammonia: 192JK -1 mol -1 Reactants = (3×131) + 192= 585 Products= (192) 2.Next, the Absolute Entropy is simply Entropy of products – Entropy of reactants. So, Products – Reactants= 585-192 =393 KJ mol -1 .
• 15.4 Spontaneity Introduction A spontaneous reaction is often defined as “a reaction that causes a system to move from a less stable to a more stable state”.  Spontaneous processes are those that can occur naturally without the need to do work. 15.4.1: Predict whether a reaction or process will be spontaneous by using the sign of ∆G  Spontaneous reactions produce substantial amounts of products at equilibrium and release free energy. o Free energy is energy that is available to do work. The Gibbs free energy change, G is the maximum amount of free energy that can be extracted to do work  If ΔG < 0, then the reaction is spontaneous  If ΔG >0, then the reaction is non-spontaneous. 15.4.2: Calculate ∆G for a reaction using the equation ∆G= ∆H- T∆S and by using values of the standard free energy change of formation, ∆G Whether a reaction is spontaneous or not depends on three factors:  Enthalpy Change (ΔH)  Entropy (ΔS)  Gibbs Free Energy (ΔG) An equation linking these three values is: ΔG = ΔH – T ΔS Gibbs Free Energy = Enthalpy Change – Temperature x Entropy Note: Temperature is measured in Kelvin, which can be calculated by 273 to the °C value you are given.  3 1 1 176 298 284 9136 0 8 91.4 1 G H T S Jmol kJmol                 3 100 176 284 620 346o G H T S T T K C           
• 15.4.3: Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and enthalpy changes and the equation ∆G= ∆H -T∆S ΔH ΔS TΔS ΔH – T ΔS Gibbs Free energy (ΔG) Spontaneous? + + + (+) – (+) Depends on “T” If ΔH < TΔS + - - (+) – (-) + Never spontaneous - + + (-) – (+) - Always spontaneous - - - (-)-(-) Depends on “T” If ΔH > TΔS 0 + + 0 – (+) - Always spontaneous 0 - - 0-(-) + Never spontaneous