Measuring Heat Of Formation Co2 Experiment Data

Introduction to Heat of Formation: Concept and Its Significance in Chemical Reactions o 5.1 Standard Heat (enthalpy) of Formation ( H f ) The superscript o signifies the standard state for the reaction at 25 o C and 1 atm. The subscript f signifies formation. The heat (or actually the enthalpy) of standard formation is often referred to as the heat of of formation. Standard heat of formation is the naming of the special enthalpy change with the formation of 1 mole of compound from its constituent elements in the standard state. An example of a picture of the enthalpy of formation of CO2 gas from its elements, namely carbon and oxygen at a temperature of 25o C and 1 atm as shown in Figure 5.1. C(s) + O2 (g) CO2 (g) Figure 5.1 Heat transfer in the steady state combustion process of C to CO at standard state lOMoARcPSD|26370826 The energy balance of the process is ∆H = Q = -393.5 kJ/g mol CO2 thus standard heat of formation of CO2 is -393.5 kJ/g mol. The standard heat of formation value for each element is zero e.g. O 2, N2 , N Heat of formation values can be found in Appendix F (Himellblau ed. 7). Standard heat of formation is negative for exothermic reactions Example 5.1 Determination of heat of formation from heat transfer measurements Determine the standard heat of formation of CO based on experimental data Completion: It is difficult to prepare pure CO from the reaction of C and O 2 and then measure the heat transfer, so it is easier to prepare the heat of formation under standard conditions for the reaction in the process shown in Figure C5.1. Figure C5.1 Q = -393.51 kJ/g mol C = ∆H A Q = -282.99 kJ/g mol CO = ∆H B a) C(s) + O2 (g) CO2 (g) b) CO (g) + ½ O2 (g) CO2 (g) Base 1 g mol C and CO Rearrange equations a) and b) to obtain equation c) as follows C (s) + ½ O2 (g) CO (g) o Hf 393.51 ( 282.99) 110.52 kJ/g mol C Example 5.2 Obtaining heat of formation from reference data What is the heat of formation of standard HCl (g)? lOMoARcPSD|26370826 Completion: o Based on Appendix F (Himellblau ed. 7), the value of H f for HCl column (g) is -92.311 kJ/g mol. Reaction of HCl formation (g) at 25 o C and 1 atm: ½ H2 (g) + ½ Cl2 (g) HCl (g) For values of H2 (g) and Cl2 (g) each has a heat of formation of o H f for column standard 0 because it is an element so the value of HCl (g) is -92.311 kJ/g mol. o Hf 1 1( 92.311) 2 1 (0) 2 (0) 92.311 kJ/g mol Problem! 1. In assessing the hazard of a compound, it can be done by based on the potential rate of energy release, one method of which is to use the heat of formation per gram of compound. Which do you think is the most explosive of the following compounds: acetylene gas, solid lead azide, liquid trinito glycerin (TNT), or solid ammonium nitrate? 2. What is the standard heat of formation of HBr (g)? 3. Show in the process that occurs in Figure 5.1 that thus an energy balance of ∆H = Q is obtained. What assumptions were used? 4. Can the heat of formation be calculated from measurements obtained from a batch process? If s o , indicate the assumptions and the calculation 5. Calculate the standard heat of formation of CH 4 based on experimental results at 25o C and 1 atm follows (Q for complete reaction) H2 (g) + ½ O2 (g) H2 O (l) C (graphite) + O2 (g) CO2 (g) CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O (l) Q = -285.84 kJ/g mol H 2 Q = -393.51 kJ/g mol C Q = -890.36 kJ/g mol CH4 Compare the calculation results obtained with the heat value formation listed in Appendix F lOMoARcPSD|26370826 o o H rxn v CH H f,CH 6 12 (1) ( 123.1) 6 12 o - v CH 66 (1) H f,CH (82.927) (3) o -vH 66 2 H f,H 2 (0) 206.0 kJ o H rxn In a complete reaction, the general equation o o products H rxn (25o C ) v H f ,i is reak tan o vi H f ,i i i (5.1) i It's important to remember that the heat of reaction is actually the enthalpy change. Example 5.3 Calculation of Standard Heat of Reciprocity based on standard heat of formation o H rxn Count For the reaction of 4 g mol NH3 shown below: 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2 O (g) Solution: Base 4 g mol NH3 o Using equation 5.1 to calculate H rxn H 25 C) for o rxn ( 4 g mol NH3 assuming a complete reaction, so : o H rxn (25 oC ) 4 (90.374) 6 ( 241.826) = -904.696 kJ/4 g mol NH3 o Per mole of reaction H rxn = -226.174 kJ/g mol NH3 What if the reaction is incomplete? ni final begin nning i Or for flow systems vi 5(0) 4( 46.191) lOMoARcPSD|26370826 out n n in v ii i So o produ k H rxn (25 C ) ( i in o reak tan enter v ) H f ,i ( i i i o v ) H f ,i i i o nmasuk H rxn (25 C ) o o H f ,i v H f ,i ii o nmasuk i H f ,i o vi H f ,i H rxn allcomponents (5.2) For example, in the reaction in Figure 5.2, the assumed conversion fraction is 0.80. C-based H66 , (0.80) (1) 1 H rxn (25 o C ) (0.80) (1) ( 123.1) ( 3 ) (0) ( 1) (82.927) 164.8 kJ Figure 5.3 Information used in the calculation H rxn (T ) lOMoARcPSD|26370826 General equation for the heat of reaction where it takes place at a temperature different from the referent: reak tan H rxn(T ) ni H i (25o C ) H i (T ) H rxn (25o C ) i products ni H i (T ) H i (25 C ) H rxn (25 C ) oo (5.3) i H (T ) H (25o C ) H(T) H(25o C ) product H rxn (25o C ) reak tan Example 5.4 Calculation of heat of reaction at different temperatures under standard conditions Reduction of CO 2 in the atmosphere is one of them by reacting the gas with hydrogen, hydrogen can be derived from electrolysis of water using solar cells. Assume a stoichiometric amount of reactants enter the reactor. Determine heat of reaction if the gas enters and leaves at 1 atm and 500 o C as shown in Figure C5.2. Figure C5.2 Completion: Base: 1 g mol CO 2 (g) at 1 atm and 500 o C Required data: lOMoARcPSD|26370826 Figure C7.3. What percentage of fuel can be saved to obtain electricity and steam by using cogeneration plants compared to separate electricity and steam plants in producing 0.25 MJ of electrical energy and 0.45 MJ of steam thermal energy. 7.3 Mechanical Energy Balance Based on Figure 7.2, the following question will arise: What size pump is used to remove biowaste from the tank, what is the pressure drop between the tank and the pipe leading to the treatment plant? Figure 7.2 Biowaste collection system The above question can be answered by using the mechanical energy balance compared to the general energy balance equation. Mechanical energy balance lOMoARcPSD|26370826 with work variables and energy i n t h e form of mechanics used for calculations involving gas compression and liquid pumping processes, while in processes such as distillation columns or reactors, the important energy is in the form of heat transfer and enthalpy change, but the energy that matters is the heat transfer. mechanical form is assumed to be 0 so that the general energy balance equation can be used. Two different categories of energy quality: a) Energy in mechanical form such as kinetic energy, potential energy, and work. b) Energy other than in part (a) such as internal energy and heat Steady-state mechanical energy balance for systems where there is mass change with the environment (basis per unit mass) P2 (EK EP ) P1 V dp W Ev 0 (7.1) Where Ev i s the mechanical energy lost, equation 7.1 is called Bernoulli equation for reverberant = 0. If the fluid processes where Ev is incompressible, and W = 0 then p 2 V p g h 0 (7.2) 2 Mechanical energy balance is best applied to fluid flow calculations when kinetic and potential energy and work are important parts. Friction loss can be evaluated based on friction factors or orifice coefficients. Example 7.4 Comparison Reversible work for a batch process where the process operating under the same conditions. Five cubic feet of ideal gas at 100o F will be compressed adiabatically from 1 atm to 10 atm. The equation of state used for the gas is pV1.40 = constant. There are 2 options for the compression process: lOMoARcPSD|26370826 a) Compression in a horizontal cylinder with a piston. b) Compression in rotary compressors. Calculate the reversible work required for each of these processes. Solution: 1. Process (a) of compression in a horizontal cylinder with piston The process takes place in a closed system as shown in Figure 7.1 but the cylinder is horizontal, the energy balance equation becomes ∆E = ∆U = W If any value of Cv exists, then for the value of W can be obtained by integrating Cv dT but in this case the value of Cv or C p is not given. The value of W can be calculated in another way, namely by integrating the pdV for the reversible work process Stage 1,2,3, and 4 V value1 is known, V value1is calculated. The system is a gas. V2 V p 1 / 1.40 1 5 1 1 1/1.40 0.965 ft 3 10 p2 Stage 5 Base: 5 ft3 gas at 100o F and 1 atm. Stage 6-9 V2 V 2 0.965 Wrev V1 5 p dV V1 V 1 p V 1.40 1.40 dV p 1 V1 V2 V1 1.40 V dV 2 p1 V 11.40 1 1.40 (V2 0.40 V 40 0. ) pV 1 2 2 1 pV 1 1 1.40 (10)(0.965) (1)(5) (ft3 )(atm) 1,987 Btu 0.40 0.7302 ( ft3 )(atm) 31.6 Btu (A positive sign indicates that work is performed within the system) lOMoARcPSD|26370826 2. Process (b) compression on a rotary compressor. (EK P2 EP ) V dp P1 W Ev 0 Figure C7.4 The values of EK and EP are assumed to be 0, and Ev = 0, thus P2 W P1 V dp Stage 1,2,3, and 4 As per figure C7.4 showing the system and data. So the moles of gas are p1 n1 1atm 5 ft. 3 V1 1(lbmol) (o R) 560o RT1 ( ft 3 R ) (atm) .0122 lbmol 0 Stage 5 Basis = 0.0122 lb mol Stage 6,7,8, and 9 P2 P2 W rev V dp V P1 Vp W rev n1 1 . 350 p 0.286 n1 1 dp p 2 V 1 / 1.4 1 P1 0.714 W rev p -p p 1 0.286 0.714 1 0.286 3 .50 p2 0.286 - p1 n1 = (5) (1)0.714 [(3.50) (10)0.286 - 10.286 ] (ft3 ) (atm) (1.987 Btu/0.7302 (ft ) 3 (atm)) = 44.4 Btu lOMoARcPSD|26370826 Problem! 1. Water is pumped from a large water tank, as shown in figure C7.5 at a rate of 2000 gal/min. Determine the minimum power (e.g. in reversible processes) that the pump requires in horsepower. Figure C7.5 2. Use the mechanical energy balance to explain why the flow of water down a vertical pipe filled with water is different from the flow down a waterfall. 3. Fluid with a specific gravity of 1.15 is discharged from the bottom of an open tank into a container through a 5 cm diameter (inside diameter) pipe at a height of 4.5 m below the fluid level of the tank. Calculate the discharge velocity of the fluid, ignoring the effects of friction. lOMoARcPSD|26370826 Figure 9.7 Graph of humidity (psychometric) in AE units lOMoARcPSD|26370826 Figure 9.8 Graph of humidity (psychometric) in SI units lOMoARcPSD|26370826 Example 9.1 Determining Wet Air Properties with Humidity Graphs Write down all the properties that you can find in the humidity (AE units) graph for wet air at a dry-bulb temperature of 90o F and a wet-bulb temperature of 70o F. It can be assumed that one can measure dry-bulb temperature using a mercury thermometer and wet-bulb temperature using a sling. psychrometer. Solution: Completion by looking at the AE unit humidity graph. Chart explanation: Determine point A of dry-bulb temperature (90o F) and wet-bulb temperature (70 o F) by drawing a line upward from T DB = 90o F then intersecting with the wet-bulb line at 70o F and drawing it to the left until it reaches saturated air line (100% relative humidity). From point A, it is drawn to the left until the humidity line (H). Other properties can be determined after point A is fixed, such properties include : lOMoARcPSD|26370826 Two models were used to determine the number of variables associated with the process flow N sp is the number of components in the flow. The calculation of the number of compositions uses N sp - 1 instead of N sp because the implicit constraint on the number of mass fractions or moles is 1. So it can be concluded that the number of variables Nv needed to determine the flow conditions in a straightforward manner are Nv = N sp + 2 Some types of equations in degree of freedom analysis: 1. Independent mass balance for each component (total balance can be substituted for 1 component balance) 2. Energy balance 3. Phase equilibrium relationship 4. Chemical equilibrium relationships 5. Implicit relationships such as component concentrations are zero in flow 6. Explicit relations such as the fraction given in flow condensation 7. Reaction rate or fraction conversion. Here's an example of calculating degrees of freedom Isobaric, steady-state, isothermal process where there are three flows plus heat transfer. The calculation of variables and equations is as follows: Variables: (N sp + 2) x 3 = (2 + 2) x 3 =12 lOMoARcPSD|26370826 Q 1 Total 13 Equation: Mass Balance 2 Energy Balance 1 H equilibrium2 O 1 T is the same in 3 streams (T1 = T2 = T3) 2 independent equations 2 p is the same in 3 streams (p1 = p2 = p3 ) 2 independent equations Total 2 8 Degrees of freedom: 13 - 8 =5 lOMoARcPSD|26370826 Figure 10.1 A simple process with three flows. In figure A, the flow is the mole flow rate and in figure B the flow is the total flow and the component composition is shown in mole fraction. Example Problem 10.1 An isothermal separation column is shown in figure C10.1. Specifications column for incoming material (feed)with the following mass fractions: 0.15, C 5 overhead = 0.20, C The followin C g: 5 4 C i5 = 0.40 and = 0.30, C 6 and C 6 C 4 = = 0.35. Mass fraction of the part = 0. In the residual product, the mass fraction as = 0. Figure C10.1 If F = 100 lb/hr is known, can the separation column be fully specified where the degrees of freedom Nd = 0? The flows P 1 and P2 are not in equilibrium. Completion: Step one: calculate Nv and Ne . We will assume that all flow temperatures and pressures are identical. Total Nv : Nv = ( N sp + 2) (3) = (4 + 2) (3) = 18 Sum of constraints and equations (Ne ): Components of mass number (no reaction) = N 4 sp = TF = TP1 = TP2 2 p F= p P1= p P2 2 lOMoARcPSD|26370826 Initial column specifications: (4 in F, 2 in P1 , and 1 in P1 plus T F = 30o C and p F = 1 atm) 9 Total N e 17 Number of degrees of freedom Nd : 18 - 17 = Problem! 1. Determine the degrees of freedom for the condenser as shown in the figure below 2. Determine the degrees of freedom for the reboiler as shown below. What variables must be specified to create the energy balance and mass balance solutions? 3. If at the stage equilibrium stage as shown below, feed flow is added. Determine the degrees of freedom in this case. 1