Enthalpy

Enthalpy For chemical reactions, the change in enthalpy (ΔH) is equal to the amount of heat absorbed or released at constant pressure. Example: C(s) + O2(g)  CO2(g) ΔH = −393.5 kJ If ΔH is positive, the reaction absorbs heat (endothermic) If ΔH is negative, the reaction releases heat (exothermic)

Enthalpy For certain reactions, measuring this change in enthalpy may be difficult (especially if the reaction proceeds too slowly, or is part of a series of reactions). Hess’s Law states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps.

Enthalpy Example – Given the two standard enthalpies below, C(s) + O2(g)  CO2(g) ΔH = −393 kJ CO2(g)  CO(g) + ½O2(g) ΔH = 283 kJ What is the reaction heat for the chemical reaction: C(s) + ½O2(g)  CO(g)? Add the two equations and cancel out like terms. C(s) + O2(g)  CO2(g) ΔH = −393 kJ CO2(g)  CO(g) + ½O2(g) ΔH = 283 kJ C(s) + O2(g) + CO2(g)  CO2(g) + CO(g) + ½O2(g) Only ½O2 remains

Enthalpy Example – Given the two standard enthalpies below, C(s) + O2(g)  CO2(g) ΔH = −393 kJ CO2(g)  CO(g) + ½O2(g) ΔH = 283 kJ What is the reaction heat for the chemical reaction: C(s) + ½O2(g)  CO(g)? Add the two equations and cancel out like terms. C(s) + O2(g)  CO2(g) ΔH = −393 kJ CO2(g)  CO(g) + ½O2(g) ΔH = 283 kJ C(s) + ½O2(g)  CO(g) ΔH = −110 kJ

Enthalpy For many problems, you will need to reverse one of the standard enthalpy equations. When this is done, the sign of ΔH switches. Example: C(s) + O2(g)  CO2(g) ΔH = −393 kJ CO2(g)  C(s) + O2(g) ΔH = 393 kJ

Enthalpy It is also likely that you will need to multiply one or both of the standard enthalpy equations by some factor. When this is done, ΔH is also increased by this factor. Example: 2[C(s) + O2(g)  CO2(g) ΔH = −393 kJ] 2C(s) + 2O2(g)  2CO2(g) ΔH = −786 kJ

Enthalpy For these types of problems, look for substances that are unique to one of the standard enthalpy equations. These substances need to be on the same side of the equation and have the same coefficient as the overall chemical reaction. Example: 2Al(s) + O2(g)  Al2O3(s) ΔH = −1,670 kJ 2Fe(s) + O2(g)  Fe2O3(s) ΔH = −842 kJ Reaction – 2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) 3 2 3 2 Same side, same coefficient On other side! Reverse this equation.

Enthalpy Unique substances Example – Given the two equations below, PCl5(s)  PCl3(g) + Cl2(g) ΔH = 88 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ What is the enthalpy change for the chemical reaction: 2P(s) + 5Cl2(g)  2PCl5(s)? Different side, Different coefficient Same side, same coefficient *Must reverse the first equation and multiply it by 2.

Enthalpy Example – Given the two equations below, PCl5(s)  PCl3(g) + Cl2(g) ΔH = 88 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ What is the enthalpy change for the chemical reaction: 2P(s) + 5Cl2(g)  2PCl5(s)? 2[PCl3(g) + Cl2(g)  PCl5(s) ΔH = −88 kJ] PCl5(s) PCl3(g) + Cl2(g) − 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ

Enthalpy Example – Given the two equations below, PCl5(s)  PCl3(g) + Cl2(g) ΔH = 88 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ What is the enthalpy change for the chemical reaction: 2P(s) + 5Cl2(g)  2PCl5(s)? 2[PCl3(g) + Cl2(g)  PCl5(s) ΔH = −88 kJ] 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ

Enthalpy Example – Given the two equations below, PCl5(s)  PCl3(g) + Cl2(g) ΔH = 88 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ What is the enthalpy change for the chemical reaction: 2P(s) + 5Cl2(g)  2PCl5(s)? 2PCl3(g) + 2Cl2(g)  2PCl5(s) ΔH = −176 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ

Enthalpy Example – Given the two equations below, PCl5(s)  PCl3(g) + Cl2(g) ΔH = 88 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ What is the enthalpy change for the chemical reaction: 2P(s) + 5Cl2(g)  2PCl5(s)? 2PCl3(g) + 2Cl2(g)  2PCl5(s) ΔH = −176 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ 2PCl3(g) + 2Cl2(g) + 2P(s) + 3Cl2(g)  2PCl5(s) + 2PCl3(g) Combine

Enthalpy Example – Given the two equations below, PCl5(s)  PCl3(g) + Cl2(g) ΔH = 88 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ What is the enthalpy change for the chemical reaction: 2P(s) + 5Cl2(g)  2PCl5(s)? 2PCl3(g) + 2Cl2(g)  2PCl5(s) ΔH = −176 kJ 2P(s) + 3Cl2(g)  2PCl3(g) ΔH = −574 kJ 2P(s) + 5Cl2(g)  2PCl5(s) ΔH = −750 kJ

Enthalpy 1. Which equation(s) below need(s) to be reversed for the reaction N2(g) + O2(g)  2NO(g)? I. 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l) II. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l) a) I. only b) II. only c) both d) neither

Enthalpy 2. Which equation(s) below need(s) to be multiplied by a factor for the reaction N2(g) + O2(g)  2NO(g)? I. 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l) II. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l) a) I. only b) II. only c) both d) neither

Enthalpy 3. What is the reaction heat for N2(g) + O2(g)  2NO(g) given the enthalpy changes below? I. 4NH3(g) + 3O2(g)  2N2(g) + 6H2O(l) ΔH = -1,530 kJ II. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l) ΔH = -1,170 kJ a) -360 kJ b) -180 kJ c) 360 kJ d) 180 kJ

Enthalpy 4. What can you tell about the reaction below based on the change in enthalpy? H2O(g) + CH4(g)  CO(g) + 3H2(g) ΔH = 206 kJ a) The reaction is exothermic b) The reaction is endothermic c) The reaction is spontaneous d) The reaction is reversible