# 5.7 Narrated Notes

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### Chapter 5.7 – Enthalpies of Formation :

Chapter 5.7 – Enthalpies of Formation

### Enthalpies of Formation :

We can calculate the enthalpy changes for a great many reactions from tabulated ΔH values 5.7 Enthalpies of Formation

### Enthalpy of Formation :

The enthalpy change associated with the formation of a compound from its constituent elements is called the enthalpy of formation 5.7 Enthalpy of Formation

### Enthalpy of Formation :

The magnitude of any enthalpy change depends on the conditions of temperature, pressure, and state 5.7 Enthalpy of Formation

### Enthalpies of Formation :

In order to compare the enthalpies of different reactions, we must define a set of conditions The set of standard conditions is called the standard state 5.7 Enthalpies of Formation

### Standard State :

What are the standard state conditions? Pressure = 1 atm Temperature = 298 K 5.7 Standard State

### Standard Enthalpy Change :

The standard enthalpy change of a reaction is defined as the enthalpy change when all reactants and products are in their standard states 5.7 Standard Enthalpy Change

### Standard Enthalpy of Formation :

The standard enthalpy of formation of a compound is the change in enthalpy for the reaction that forms one mole of the compound from its elements, with all substance in their standard state 5.7 Standard Enthalpy of Formation

### Standard Enthalpy of Formation :

The most stable form of the element is usually used for the formation reaction What is the most stable form of hydrogen? H2(g) 5.7 Standard Enthalpy of Formation

### Standard Enthalpy of Formation :

The stoichiometry of formation reactions always indicates that one mole of the desired substance is produced This means we can use fractions as coefficients for the reactants 5.7 Standard Enthalpy of Formation

### Standard Enthalpy of Formation :

Why is the standard enthalpy of formation of the most stable form of any element zero? Because there is no formation reaction needed when the element is already in its standard state 5.7 Standard Enthalpy of Formation

### Practice Exercise :

Write the equation corresponding to the standard enthalpy formation of liquid carbon tetrachloride C(s) + 2 Cl2(g) CCl4(l) 5.7 Practice Exercise

### Using Enthalpies of Formation to Calculate Enthalpies of Reaction :

We can use Hess’s law to calculate the standard enthalpy change for any reaction for which we know the ΔH˚fvalues for all reactants and products Equation [5.31] = ΣnΔH˚f(products) - ΣmΔH˚f (reactants) 5.7 Using Enthalpies of Formation to Calculate Enthalpies of Reaction

### Practice Exercise :

Using the standard enthalpies of formation listed in Table 5.3, calculate the enthalpy change for the combustion of 1 mol of ethanol: C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l) ∆H =[(2⋅-393.5) + (3⋅-285.83)] – [-277.7] = -1367 kJ 5.7 Practice Exercise

### Practice Exercise :

Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s) CuO(s) + H2(g)Cu(s) + H2O(l)ΔH˚ = -129.7 kJ -129.7 kJ = [0 + -285.83] – [(x) + 0] x= -156.1 kJ/mol 5.7 Practice Exercise