CHAPTER 4 HEAT EFFECT

Consider the process of manufacturing ETHYLENE GLYCOL (an antifreeze agent) from ethylene : -Vaporization -Heating Ethylene (liquid) (-30 o C ?) Ethylene + O 2  ethylene oxide  +H 2 O (vapor) (gas) (gaseous) ETHYLENE GLYCOL -Reaction -Cooling -Reaction -Mixing (dissolution) -Condensation -Cooling/Heating

Approximate true

Example 4-1: What is the final temperature when 4x10 5 Btu are added to 25 lbmole of ammonia, initially at 500 o F in a steady process at approximately 1 atm?

The heat capacity of the gas MIXTURE in the ideal gas state is given by: C p ig mixture = y A C p ig A + y B C p ig B +y C C p ig C + … = ∑y i C p ig i Where : C p ig A, C p ig B, C p ig i = heat capacity of gas A, B, and i in ideal gas state, respectively. y A, y B, y i = mole fraction of gas A, B, and I, redpectively.

4.2. STANDART HEAT OF REACTION Heat of reaction occurs due to changes in molecular structure. Differences in molecular structure  differences in energy content. STANDARD HEAT : Consider reaction : aA + bB  l L + mM Standard heat of reaction : enthalpy change when a mole of A and b mole of B in their standard states at temperature T reacts to form l moles of L and m moles of M in their standard states, also at temperature T.

Standard heat of formation Heat of formation : the heat effect resulted from a formation reaction, based on 1 mole of the compound formed. Formation reaction : a reaction which forms a single compound from its constituent elements. Example : (a) C + ½ O 2 + 2H 2  CH 3 OH(formation reaction) (b) H2O + SO3  H2SO4(NOT a formation reaction) Same examples of heat of formation : CO 2(g) :C (s) + O 2(g)  CO 2(g) ∆H f o 298 = J H 2 (g) :since H 2 is an element ∆H f o 298 = 0 CO (g) :C (s) + ½ O 2(g)  CO (g) ∆H f o 298 = J H 2 O (g) :H 2(g) +1/2 O 2(g)  H 2 O (g) ∆H f o 298 = J

In many cases, heat of reaction can be calculated from heat of formation, e.g. CO 2(g) + H 2(g)  CO (g) + H 2 O (g) ∆H f o 298 = ? CO 2(g)  C (s) + O 2(g) ∆H f o 298 = J C (s) + ½ O 2(g)  CO (g) ∆H f o 298 = J H 2(g) + ½ O 2(g)  H 2 O (g) ∆H f o 298 = J + CO 2(g) + H 2(g)  CO (g) + H 2 O (g) ∆H f o 298 = J

Standard heat of combustion A combustion reaction is defined as a reaction between an element or compound and oxygen to form specified combustion products. In some cases, heat of formation can be calculated from heat of combustion (which is measurable), e.g. 4C (s) + 5H 2(g)  C 4 H 10(g) 4C (s) + 4O 2(g)  4CO 2(g) ∆H f o 298 = 4( ) 5H 2(g) + 2½ O 2(g)  5H 2 O (l) ∆H f o 298 = 5( ) 4CO 2(g) + 5H 2 O (l)  C 4 H 10(g) + 6½ O2(g) ∆H f o 298 = C (s) + 5H 2(g)  C 4 H 10(g) ∆H f o 298 = J

TEMPERATURE DEPENDENCE OF ∆H o the general chemical reaction may be written as: lv 1 lA 1 + lv 2 lA 2 + …  lv 3 lA 3 + lv 4 lA 4 + … Where :lv 1 l = stoichiometric coefficient of Ai lv 1 l =+ for products - for reactants Example :N 2 + 3H 2  2 NH 3 v N2 = -1 ; v H2 = -3 ; v NH3 = +2 The std. rate of rxn : ∆H ≡ ∑ v i. H i o H i o = [enthalpy of species –i in its standard state]

Example IV-1 : What is the maximum temperature that can be reached by the combustion of methane with 20% excess air? Both methane & air enter the burner at 25 o C. The reaction is : CH 4 + 2O 2  CO 2 + 2H 2 O(g) ∆H f o 298 = J

Example IV-2 : One method for the manufacture of “synthesis gas” (primarily a mixture of CO and H 2 ) is the catalytic reforming of CH 4 with steam at high temperature and atmospheric pressure : CH 4(g) + H 2 O (g)  CO (g) + 3H 2(g) The only other reaction which occurs to an appreciable extent is the water-gas-shift reaction : CO (g) + H 2 O (g)  CO 2(g) + H 2(g) If the reactants are supplied in the ratio : 2 mole steam to 1 mole CH 4, and if heat is supplied to the reactor so that the products reach a temperature of 1300 K, the CH 4 is completely converted and the product stream contains 17,4 % mole CO. assuming the reactants to be preheated to 600 K, calculate the heat requirement for the reactor. Data: CH 4(g) + H 2 O (g)  CO (g) + 3H 2(g) ∆H f o 298 = J CO (g) + H 2 O (g)  CO 2(g) + H 2(g) ∆H f o 298 = J