Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Standard enthalpy of formation and reaction.  Calculating the standard enthalpy of reaction  Direct method of calculating the standard enthalpy of formation. 1

 Standard enthalpy of formation and reaction: The standard enthalpy of formation (ΔH f ° ) is defined as the heat change that results when one mole of a compound is formed from its elements at a pressure of (1 atm  Calculating the standard enthalpy of reaction From standard enthalpies of formation, we can calculate the standard enthalpy of reaction (ΔH° rxn ). Consider the following hypothetical reaction.

aA + bB → cC + dD Where, a,b,c and d are stiochiometric coefficients. The standard enthalpy of reaction is given by: ΔH° rxn = [ c ΔH° f ( C) + d ΔH° f (D) ] – [ a ΔH° f (A) + b ΔH° f (B)] Where a,b,c and d all have the unit mol. The equation can be written in the general form: ΔH° rxn = ( (Σn ΔH° f (products) –( Σm ΔH° f (reactnts) )  Where (m) and (n) denote the stiochiometric coefficients for the reactants and products,and ( Σ ) (sigma) means '' the sum of''.

Example: A reaction used for rocket engines is N 2 H 4 (l) + 2H 2 O 2 (l) → N 2 (g) +4H 2 O (l) What is the standard enthalpy of reaction in kilojoules ? the standard enthalpies of formation are: ΔH° f [ N 2 H 4 (l)] = 95.1 kJ. ΔH° f [ H 2 O 2 (l)] = kJ. ΔH° f [ H 2 O(l) ] = kJ. Solution : ΔH° rxn = ( (Σn ΔH° f (products) –( Σm ΔH° f (reactnts) ) ΔH° rxn = ΔH° f [N 2 (g) ] +4ΔH° f [H 2 O(l)]-{ ΔH° f [N 2 H 4 (l)]+2[H 2 O 2 (l)]}

 Remember, the standard enthalpy of formation of any element in its most stable form is zero. Therefore, ΔH° f [ N 2 (g)] = 0. ΔH° rxn = [ 0 + 4( kJ)] –[ 95.1 kJ+ 2( kJ)] = kJ.  Direct method of calculating the standard enthalpy of formation  This method of measuring (ΔH° f ) applies to compounds that can be readily synthesized from their elements. The best way to describe this direct method is to look at the following example.

Example: The combustion of sulfur occurs according to the following thermo chemical equation: S (rhombic) + O 2 (g) → SO 2 (g) ΔH° rxn = kJ. What is the enthalpy of formation of (SO 2 (g)). Solution: Step(1): recall that the enthalpy of reaction carried out under standard-state conditions is given by ΔH° rxn = ( (Σn ΔH° f (products) –( Σm ΔH° f (reactants) ) ΔH° rxn = [ΔH° f (SO 2 )(g) ] – [ΔH° f (S) + ΔH° f (O 2 ) (g) ]

Step(2): recall that the standard enthalpy of formation of any element in its most stable form is zero. Therefore, ΔH° f of (S) =0 and ΔH° f of (O 2 ) (g) = 0. ΔH° rxn = [ΔH° f (SO 2 )(g) ] – [ΔH° f (S) + ΔH° f (O 2 ) (g) ] -296 kJ = [ΔH° f (SO 2 )(g) ] –[ 0+0] ΔH° f (SO 2 )(g) = -296 kJ/ mol (SO 2 ). Note: you should recognize that this chemical equation as written meets the definition of a formation reaction. Thus, ΔH° rxn is ΔH° f of (SO 2 ) (g).

 Indirect method of calculating the standard enthalpy of formation, (Hess's law):  Hess's law, states that when reactants are converted to products, the change in enthalpy is the same wither the reaction takes place in one step on in a series of steps.  This means that if we can break down the reaction of interest into a series of reactions for which (ΔH° rxn ) can be measured, we can calculate (ΔH° rxn ) for the overall reaction. Let's look at an example: Example: For the following heats of combustion with fluorine, calculate the enthalpy of formation of methane, (CH 4 ):

(a) CH 4 (g) + 4F 2 (g) → CF 4 (g) + 4HF (g) ΔH° rxn = kJ. (b) C(graphite) + 2F 2 (g) → CF 4 (g) ΔH° rxn = kJ. ( c ) H 2 (g) + F 2 (g) → 2 HF (g) ΔH° rxn = kJ. Solution: Step (1): the enthalpy of formation of methane can be determined from the following equation. C(graphite) + 2H 2 (g) → CH 4 (g) ΔH° rxn = ?  If we can calculate (ΔH° rxn ) for this reaction, we can calculate ΔH° f (CH 4 ) because the enthalpies of formation of ( C ) and ( H 2 (g) ) are zero. Why ? Remember that the standard enthalpy of formation of any element in its most stable form is zero.

Step (2): since we want to obtain one equation containing only (C,H 2, and CH 4 ), we need to eliminate ( F 2, CF 4, and HF), from the first three equation (a-c). first, we note that equation (a) contains methane (CH 4 ), on the reactant side. Let's reverse (a) to get (CH 4 ) on the product side. CF 4 (g) + 4HF (g) →CH 4 (g) + 4F 2 (g) ΔH° rxn = kJ. Note : ΔH° rxn changed sign when reversing the direction of the reaction.

CF 4 (g) + 4HF (g) → CH 4 (g) + 4F 2 (g) ΔH° rxn = kJ. C(graphite) + 2F 2 (g) → CF 4 (g) ΔH° rxn = kJ. 2H 2 (g) + 2F 2 (g) → 4HF (g) ΔH° rxn = kJ. C(graphite) + 2H 2 (g) → CH 4 (g) ΔH° rxn = - 75 kJ. Step (3): since the above equation represents the synthesis of (CH 4 ) from its elements, we have.

ΔH° rxn = [ΔH° f (CH 4 )] – [ΔH° f (C) + 2ΔH° f (H 2 )] -75 kJ = [ΔH° f (CH 4 )] – [0 + 0] ΔH° f (CH 4 ) = -75 kJ / mol (CH 4 ).  The first law of thermodynamics:  Applying the first law of thermodynamics:  The first law of thermodynamics states that energy can be converted form one form to another, but cannot be created or destroyed. ΔE sys + ΔE surr = 0

 Where: The subscripts '' sys'' and '' surr'' denote system and surroundings, respectively. ΔE = q + w Where: ΔE is the change in internal energy of the system. q is the heat exchange between the system and surroundings. w is the work done on (or by ) the system.

 Using the sign convention for thermo chemical processes, (q is positive) for an endothermic process and (q is negative) for an exothermic process. For work, (w is positive ) for (work done on the system by the surroundings) and (w is negative for work done by the system on the surroundings).

Example: A system does (975 kJ) of work on its surroundings while at the same time it absorb ( 625 kJ) of heat. What is the change in energy, ΔE, for the system. Solution: To solve this problem, you must make sure to get the sign convention correct. The system does work on the surroundings; this is an energy-depleting process. w = - 975kJ. The system absorbs (625 kJ) of heat. Therefore, the internal energy of the system would increase. q = + 625kJ. ΔE = q + w = 625kJ + ( - 975kJ) = kJ.