What is the difference between standard enthalpy of combustion and standard enthalpy of formation?
Q. What is the difference between standard enthalpy of combustion and standard enthalpy of formation? i have a homework problem that gives the standard enthalpy of combustion of C2H4 and wants to know what the standard enthalpy of formation is. How would i do that? Thanks
Asked by dillan - Wed Dec 17 01:03:31 2008 - - 3 Answers - 0 Comments
A. Standard enthalpy of combustion of C2H4 is the energy released when one mole of C2H2 is burned completely: C2H4 + 5/2 O2 --> 2 CO2 + H2O Standard enthalpy of formation of C2H4 is the energy needed to form one mole of C2H4 from its constituent components in standard conditions: 2 C + 2 H2 --> C2H4 This is how you answer your homework question: 2 CO2 + 2 H2O --> C2H4 + 3 O2 (invert the combustion equation) 2 C + 2 O2 --> 2 CO2 (formation of CO2) 2 H2 + O2 --> 2 H2O (formation of H2O) ___ 2 C + 2 H2 --> C2H4 Remember to reverse the sign of standard enthalpy of combustion of C2H4. You should find out the standard enthalpy of formation of CO2 and H2O. Remember to multiply those values by 2. Add all the enthalpy values you get from… [cont.]
Answered by KC - Wed Dec 17 01:56:16 2008
Q. What is the difference between standard enthalpy of combustion and standard enthalpy of formation? i have a homework problem that gives the standard enthalpy of combustion of C2H4 and wants to know what the standard enthalpy of formation is. How would i do that? Thanks
Asked by dillan - Wed Dec 17 01:03:31 2008 - - 3 Answers - 0 Comments
A. Standard enthalpy of combustion of C2H4 is the energy released when one mole of C2H2 is burned completely: C2H4 + 5/2 O2 --> 2 CO2 + H2O Standard enthalpy of formation of C2H4 is the energy needed to form one mole of C2H4 from its constituent components in standard conditions: 2 C + 2 H2 --> C2H4 This is how you answer your homework question: 2 CO2 + 2 H2O --> C2H4 + 3 O2 (invert the combustion equation) 2 C + 2 O2 --> 2 CO2 (formation of CO2) 2 H2 + O2 --> 2 H2O (formation of H2O) ___ 2 C + 2 H2 --> C2H4 Remember to reverse the sign of standard enthalpy of combustion of C2H4. You should find out the standard enthalpy of formation of CO2 and H2O. Remember to multiply those values by 2. Add all the enthalpy values you get from… [cont.]
Answered by KC - Wed Dec 17 01:56:16 2008
Explain why the standard enthalpy of formation should be different for propanal and propanone?
Q. Should the answer be because they have different chemical properties and structural formula? I asked the question since by working a theoretical value for the standard enthalpy of formation of both propanone and propanal by calculating their bond energy terms will result in the same answer, but in reality they should be different.
Asked by Jonathan24 - Wed Apr 23 06:23:20 2008 - - 1 Answers - 0 Comments
A. they are different compounds, & they likelihood of two entirely different compounds having the same stability... is mighty slim using bond energies, uses the "average energy" of a bond ... and so only gives you a hint as to what the energetics of a compound might actually be. in real life some compounds with a given bonding are more stable (less energetic) than your average compound, while some are the reverse... even though they may all have the same bonding for example the typical C-C bond in a SP3 hybrid shoud be about 109 degrees away from the other bonds... but if it was cyclic it might be stressed into about a 90 degree angle as in "cubane"... ring stress = instability = more energetics
Answered by Steve O - Wed Apr 23 11:58:28 2008
Q. Should the answer be because they have different chemical properties and structural formula? I asked the question since by working a theoretical value for the standard enthalpy of formation of both propanone and propanal by calculating their bond energy terms will result in the same answer, but in reality they should be different.
Asked by Jonathan24 - Wed Apr 23 06:23:20 2008 - - 1 Answers - 0 Comments
A. they are different compounds, & they likelihood of two entirely different compounds having the same stability... is mighty slim using bond energies, uses the "average energy" of a bond ... and so only gives you a hint as to what the energetics of a compound might actually be. in real life some compounds with a given bonding are more stable (less energetic) than your average compound, while some are the reverse... even though they may all have the same bonding for example the typical C-C bond in a SP3 hybrid shoud be about 109 degrees away from the other bonds... but if it was cyclic it might be stressed into about a 90 degree angle as in "cubane"... ring stress = instability = more energetics
Answered by Steve O - Wed Apr 23 11:58:28 2008
How do you find the standard enthalpy change?
Q. How do you determine the standard enthalpy change of the reactions when you are given the following equations? For example, Ba(s) + O2 (g) + H2(g) -->Ba(OH)2 (s), and enthalpy change (delta H) = -946kJ
Asked by honey(*_*) - Thu Jan 4 21:34:19 2007 - - 2 Answers - 0 Comments
A. use this equation Hreaction = Hf (Products) - Hf (Reactants) = Sum of If you have the (delta H) of the different Chemicals then plug them into the equation.
Answered by Panda WafflesZilla - Thu Jan 4 21:39:34 2007
Q. How do you determine the standard enthalpy change of the reactions when you are given the following equations? For example, Ba(s) + O2 (g) + H2(g) -->Ba(OH)2 (s), and enthalpy change (delta H) = -946kJ
Asked by honey(*_*) - Thu Jan 4 21:34:19 2007 - - 2 Answers - 0 Comments
A. use this equation Hreaction = Hf (Products) - Hf (Reactants) = Sum of If you have the (delta H) of the different Chemicals then plug them into the equation.
Answered by Panda WafflesZilla - Thu Jan 4 21:39:34 2007
Calculate the standard enthalpy of formation for Li2O?
Q. 14.4 g of Lithium is burned in excess oxygen at constant atmospheric pressure then returned back to 25 degrees celcius. A total of 605 Kj of heat was given off by this reaction. Calculate the standard molar enthalpy of formation for Li2O. Please also explain how you got the answer so I may better understand. Thanks.
Asked by coooldogg1 - Sun Jul 12 19:41:20 2009 - - 1 Answers - 0 Comments
A. standard molar enthalpy values= ___KJ/mol, so all you need to do is calculate the number of moles for lithium and just simply divide the enthalpy by the moles Good Luck
Answered by That guy - Sun Jul 12 19:53:12 2009
Q. 14.4 g of Lithium is burned in excess oxygen at constant atmospheric pressure then returned back to 25 degrees celcius. A total of 605 Kj of heat was given off by this reaction. Calculate the standard molar enthalpy of formation for Li2O. Please also explain how you got the answer so I may better understand. Thanks.
Asked by coooldogg1 - Sun Jul 12 19:41:20 2009 - - 1 Answers - 0 Comments
A. standard molar enthalpy values= ___KJ/mol, so all you need to do is calculate the number of moles for lithium and just simply divide the enthalpy by the moles Good Luck
Answered by That guy - Sun Jul 12 19:53:12 2009
help with gen chem question about standard enthalpy of combustion?
Q. Ethanol (C2H5OH) has been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid ethanol.
Asked by Sean P - Wed Nov 12 20:54:13 2008 - - 1 Answers - 0 Comments
A. the answer is -29.8
Answered by free22 - Thu Nov 13 13:48:11 2008
Q. Ethanol (C2H5OH) has been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid ethanol.
Asked by Sean P - Wed Nov 12 20:54:13 2008 - - 1 Answers - 0 Comments
A. the answer is -29.8
Answered by free22 - Thu Nov 13 13:48:11 2008
How can standard enthalpy of combustion be under standard conditions?
Q. If standard conditions are 25C, then how can you burn something? Surely it's too hot to be standard conditions anymore?
Asked by meh - Sat Mar 7 11:22:57 2009 - - 1 Answers - 0 Comments
A. standard conditions would be the initial conditions, not the conditions going on during the process. once combustion begins, youre not at standard conditions anymore.
Answered by User 1123581321 - Sat Mar 7 11:26:52 2009
Q. If standard conditions are 25C, then how can you burn something? Surely it's too hot to be standard conditions anymore?
Asked by meh - Sat Mar 7 11:22:57 2009 - - 1 Answers - 0 Comments
A. standard conditions would be the initial conditions, not the conditions going on during the process. once combustion begins, youre not at standard conditions anymore.
Answered by User 1123581321 - Sat Mar 7 11:26:52 2009
What is the standard enthalpy for H2 and F2 (Hydrogen and Florine)?
Q. the standard enthalpy for Hydrogen Floride and the standard entlapy changes for the atomisation of its elements
Asked by Kaz Wilkosz - Mon Apr 14 04:40:12 2008 - - 1 Answers - 0 Comments
A. Standard enthalpy of formation for elements in their ground states equals zero by definition. Hence Hf (F ) = 0 Hf (H ) = 0 NIST webbook gives standard enthalpy of formation for hydrogen fluoride: Hf (HF) = -273.30 kJ/mol The enthalpy of reaction equals difference of enthalpy of formation of products minus enthalpy of formation for reactants. So enthalpy of reaction for forming a compound from elements its equal to its enthalpy of formation. But only if the product's stoichiometric factor equals one 1/2 F + 1/2 H HF => Hr = Hf (HF) - (1/2) Hf (F ) - (1/2) Hf (H ) = Hf (HF) If you write formation reaction as: F + H 2 HF you get Hr = 2 Hf (HF) - Hf (F ) - Hf (H ) = 2 Hf (HF)
Answered by schmiso - Mon Apr 14 05:28:19 2008
Q. the standard enthalpy for Hydrogen Floride and the standard entlapy changes for the atomisation of its elements
Asked by Kaz Wilkosz - Mon Apr 14 04:40:12 2008 - - 1 Answers - 0 Comments
A. Standard enthalpy of formation for elements in their ground states equals zero by definition. Hence Hf (F ) = 0 Hf (H ) = 0 NIST webbook gives standard enthalpy of formation for hydrogen fluoride: Hf (HF) = -273.30 kJ/mol The enthalpy of reaction equals difference of enthalpy of formation of products minus enthalpy of formation for reactants. So enthalpy of reaction for forming a compound from elements its equal to its enthalpy of formation. But only if the product's stoichiometric factor equals one 1/2 F + 1/2 H HF => Hr = Hf (HF) - (1/2) Hf (F ) - (1/2) Hf (H ) = Hf (HF) If you write formation reaction as: F + H 2 HF you get Hr = 2 Hf (HF) - Hf (F ) - Hf (H ) = 2 Hf (HF)
Answered by schmiso - Mon Apr 14 05:28:19 2008
How to write the equations to represent standard enthalpy of formation and combustion?
Q. I understand the basic layout, but how would you determine the change in H for each equation? Nothing is given except the elements involved... for example for formation: 1. CH3Br (l) 2. CH3COC2H5 for combustion the H value is negative, but how would we know what it is? Also, because the beginning product doesn't contain C/H/O, the results aren't always carbon dioxide and water... 1. H2 (g) 2. Al (s) Help?
Asked by Lil Bunny Foo Foo - Sat Aug 23 06:15:15 2008 - - 1 Answers - 0 Comments
A. Enthalpy of formation for CH3Br C + 3H2 + 1/2Br2 = CH3Br CH3COC2H5 3C + 4H2 + 1/2O2 = CH3COC2H5 Enthalpy of combustion H2 + 1/2O2 = H2O 4Al + 3O2 = 2Al2O3
Answered by gpina_9 - Sat Aug 23 06:57:39 2008
Q. I understand the basic layout, but how would you determine the change in H for each equation? Nothing is given except the elements involved... for example for formation: 1. CH3Br (l) 2. CH3COC2H5 for combustion the H value is negative, but how would we know what it is? Also, because the beginning product doesn't contain C/H/O, the results aren't always carbon dioxide and water... 1. H2 (g) 2. Al (s) Help?
Asked by Lil Bunny Foo Foo - Sat Aug 23 06:15:15 2008 - - 1 Answers - 0 Comments
A. Enthalpy of formation for CH3Br C + 3H2 + 1/2Br2 = CH3Br CH3COC2H5 3C + 4H2 + 1/2O2 = CH3COC2H5 Enthalpy of combustion H2 + 1/2O2 = H2O 4Al + 3O2 = 2Al2O3
Answered by gpina_9 - Sat Aug 23 06:57:39 2008
Why is the standard enthalpy of formation of O2 equal to zero?
Q. Please explain thoroughly. Thanks.
Asked by anna - Fri Dec 5 03:28:37 2008 - - 1 Answers - 0 Comments
A. Because oxygen in its standard state is O2. The definition of standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed in its standard state from its constitutent elements in their standard state at 298k and 1 atm. Since oxygen is ardy an element in its standard state, there is no standard enthalpy of formation to talk about here.
Answered by eljx2004 - Fri Dec 5 03:38:12 2008
Q. Please explain thoroughly. Thanks.
Asked by anna - Fri Dec 5 03:28:37 2008 - - 1 Answers - 0 Comments
A. Because oxygen in its standard state is O2. The definition of standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed in its standard state from its constitutent elements in their standard state at 298k and 1 atm. Since oxygen is ardy an element in its standard state, there is no standard enthalpy of formation to talk about here.
Answered by eljx2004 - Fri Dec 5 03:38:12 2008
How do I calculate the standard enthalpy of combustion per gram of C2H5OH?
Q. I know the answer's -29.8...that's not the question. I need to know HOW to calculate it. I did products minus reactants but I don't get the answer and I don't know what to do next. I saw that when someone else asked this same question...Not helpful!
Asked by KTB - Tue Nov 18 13:35:49 2008 - - 1 Answers - 0 Comments
A. Let UNITs guide you; always use them in your calculation to prevent errors Firstly, the eqn: CH3CH2OH + 7/2O2 ---> 2CO2 + 3H2O___dHox = ??kJ/mole dHox/mole = (2*dHf CO2/mole + 3*dHf H2O/mole) - dHf C2H6O/mole = ?? kJ/mole dHox/g = dHox kJ/mole * MW C2H6O g/mole Plug and SOLVE Basic mathematics is a prerequisite to chemistry I just try to help you with the methodology of solving the problem.
Answered by SciMann - Tue Nov 18 16:22:59 2008
Q. I know the answer's -29.8...that's not the question. I need to know HOW to calculate it. I did products minus reactants but I don't get the answer and I don't know what to do next. I saw that when someone else asked this same question...Not helpful!
Asked by KTB - Tue Nov 18 13:35:49 2008 - - 1 Answers - 0 Comments
A. Let UNITs guide you; always use them in your calculation to prevent errors Firstly, the eqn: CH3CH2OH + 7/2O2 ---> 2CO2 + 3H2O___dHox = ??kJ/mole dHox/mole = (2*dHf CO2/mole + 3*dHf H2O/mole) - dHf C2H6O/mole = ?? kJ/mole dHox/g = dHox kJ/mole * MW C2H6O g/mole Plug and SOLVE Basic mathematics is a prerequisite to chemistry I just try to help you with the methodology of solving the problem.
Answered by SciMann - Tue Nov 18 16:22:59 2008
How do you find standard enthalpy of formation from heat of combustion?
Q. How do you find standard enthalpy of formation from heat of combustion?
Asked by Christina T - Mon Mar 12 16:52:31 2007 - - 2 Answers - 0 Comments
A. I'll try my best to do a worked example although i prefer to draw all the arrows etc. I take it when you mean heat of combustion you mean enthalpy??? 'Enthalpy' deals entirely with energy - if you have 'heat' for something you need to work out its enthalpy of combustion first (i.e. traditionally temperature raised in a set volume of water, Q=mc T). Calculate Hf of propan-1-ol given the following Hc: Propan-1-ol = -2010 C = -394 H2 = -298 3C + 4H2 --->C2H5CH2OH (Eqn for Hf) Combustion products of above: 3CO2 + 4H20 (Arrows would point down to these from either side of our Hf equation). Arrows pointing in the same direction go on the same side of the equation, thus our formula (remember the moles): Hf + Hc C2H5CH2OH = [(3X Hc C)+(4 [cont.]
Answered by Antimonic - Mon Mar 12 17:47:00 2007
Q. How do you find standard enthalpy of formation from heat of combustion?
Asked by Christina T - Mon Mar 12 16:52:31 2007 - - 2 Answers - 0 Comments
A. I'll try my best to do a worked example although i prefer to draw all the arrows etc. I take it when you mean heat of combustion you mean enthalpy??? 'Enthalpy' deals entirely with energy - if you have 'heat' for something you need to work out its enthalpy of combustion first (i.e. traditionally temperature raised in a set volume of water, Q=mc T). Calculate Hf of propan-1-ol given the following Hc: Propan-1-ol = -2010 C = -394 H2 = -298 3C + 4H2 --->C2H5CH2OH (Eqn for Hf) Combustion products of above: 3CO2 + 4H20 (Arrows would point down to these from either side of our Hf equation). Arrows pointing in the same direction go on the same side of the equation, thus our formula (remember the moles): Hf + Hc C2H5CH2OH = [(3X Hc C)+(4 [cont.]
Answered by Antimonic - Mon Mar 12 17:47:00 2007
What is standard enthalpy for the following chemical reaction?
Q. CO2(g)+ 2 KOH(s) = H2O(g)+ K2CO3(s)
Asked by Derek - Tue Oct 14 18:07:14 2008 - - 1 Answers - 0 Comments
A. Sum the enthalpies of formation of the products and subtract the enthalpies of the reactants. Make sure you multiply each substance's enthalpy by its stoichiometric coefficient. The standard enthalpies of formation should be on a table in your book.
Answered by Joe - Tue Oct 14 18:15:49 2008
Q. CO2(g)+ 2 KOH(s) = H2O(g)+ K2CO3(s)
Asked by Derek - Tue Oct 14 18:07:14 2008 - - 1 Answers - 0 Comments
A. Sum the enthalpies of formation of the products and subtract the enthalpies of the reactants. Make sure you multiply each substance's enthalpy by its stoichiometric coefficient. The standard enthalpies of formation should be on a table in your book.
Answered by Joe - Tue Oct 14 18:15:49 2008
What is the equation that represents the standard enthalpy of formation for KClO3?
Q. What is the equation that represents the standard enthalpy of formation for KClO3?
Asked by dco123 - Mon Mar 17 01:41:13 2008 - - 1 Answers - 0 Comments
A. kJ/mol
Answered by Jay - Mon Mar 17 01:52:05 2008
Q. What is the equation that represents the standard enthalpy of formation for KClO3?
Asked by dco123 - Mon Mar 17 01:41:13 2008 - - 1 Answers - 0 Comments
A. kJ/mol
Answered by Jay - Mon Mar 17 01:52:05 2008
How do i calculate the standard enthalpy of combustion per gram of liquid ethanol?
Q. the formula is C2H5OH
Asked by doggy - Tue Nov 11 16:35:41 2008 - - 2 Answers - 0 Comments
A. i got the answer... it's -29.8
Answered by free22 - Thu Nov 13 13:43:57 2008
Q. the formula is C2H5OH
Asked by doggy - Tue Nov 11 16:35:41 2008 - - 2 Answers - 0 Comments
A. i got the answer... it's -29.8
Answered by free22 - Thu Nov 13 13:43:57 2008
Standard Enthalpy Change of these 5 alcohols?
Q. I can't seem to find this on the internet. I need to know the standard enthalpy change of combustion of: Methanol Ethanol Propan-1-ol Propan-2-ol Butan-1-ol
Asked by haydenriach - Fri Dec 14 18:14:12 2007 - - 1 Answers - 1 Comments
A.
Answered by neoga illinois - Fri Dec 14 19:46:18 2007
Q. I can't seem to find this on the internet. I need to know the standard enthalpy change of combustion of: Methanol Ethanol Propan-1-ol Propan-2-ol Butan-1-ol
Asked by haydenriach - Fri Dec 14 18:14:12 2007 - - 1 Answers - 1 Comments
A.
Answered by neoga illinois - Fri Dec 14 19:46:18 2007
Which of the following standard enthalpy of formation values are not zero at 25 C?
Q. Br2(l) Si(s) S8(s)(rhombic) H2(g) Br2(l) I thought it was H2 in gram. how do i know??? SiO2(s)
Asked by Jenna - Sat Jun 28 17:33:28 2008 - - 2 Answers - 0 Comments
A. Substances have a stand enthalpy of formation of 0 when they are in their standard state, as the first poster said. This is the physical state (solid, liquid, or gas) that a substance would be in under standard conditions. Thus, the standard state for carbon is solid, for water is liquid and for hydrogen is gas. Why? Because at 1.00 atm. and 25 C, these substances are in the physical state specified. I'm not sure why you have 2 Br2(l)'s but if one of them is in gas (g) phase, that won't have a enthalpy of formation of 0 then. Si(s), S8(s) [rhombic], H2(g) are all 0kJ/mol. Note that (g) does not mean GRAMS... it indicates the GASEOUS phase. Again, as the first poster said, if SiO2(s) is part of the list, then it must be that, since… [cont.]
Answered by /\/ ? - Sat Jun 28 22:30:46 2008
Q. Br2(l) Si(s) S8(s)(rhombic) H2(g) Br2(l) I thought it was H2 in gram. how do i know??? SiO2(s)
Asked by Jenna - Sat Jun 28 17:33:28 2008 - - 2 Answers - 0 Comments
A. Substances have a stand enthalpy of formation of 0 when they are in their standard state, as the first poster said. This is the physical state (solid, liquid, or gas) that a substance would be in under standard conditions. Thus, the standard state for carbon is solid, for water is liquid and for hydrogen is gas. Why? Because at 1.00 atm. and 25 C, these substances are in the physical state specified. I'm not sure why you have 2 Br2(l)'s but if one of them is in gas (g) phase, that won't have a enthalpy of formation of 0 then. Si(s), S8(s) [rhombic], H2(g) are all 0kJ/mol. Note that (g) does not mean GRAMS... it indicates the GASEOUS phase. Again, as the first poster said, if SiO2(s) is part of the list, then it must be that, since… [cont.]
Answered by /\/ ? - Sat Jun 28 22:30:46 2008
Which of the following does not have a standard enthalpy of formation equal to zero at 25 C and 1.0 atm..?
Q. Which of the following does not have a standard enthalpy of formation equal to zero at 25 C and 1.0 atm? A) F2(g) B) Al(s) C) H2O(l) D) H2(g) E) They all have a standard enthalpy equal to zero.
Asked by the beast - Sun Jun 7 17:51:49 2009 - - 1 Answers - 0 Comments
Q. Which of the following does not have a standard enthalpy of formation equal to zero at 25 C and 1.0 atm? A) F2(g) B) Al(s) C) H2O(l) D) H2(g) E) They all have a standard enthalpy equal to zero.
Asked by the beast - Sun Jun 7 17:51:49 2009 - - 1 Answers - 0 Comments
Calculating the standard enthalpy of a combustion?
Q. The standard enthalpy of combustion of ethene gas, C2H4(g), is -1411.1 kJ/mol at 298 K. Given the following enthalpies of formation, calculate Hf for C2H4(g). CO2(g) -393.5 kJ/mol H2O(l)-285.9 kJ/mol
Asked by NoName - Sat Nov 8 15:39:20 2008 - - 1 Answers - 0 Comments
A. Firs write the balanced equation: C2H4 + 3 O2 ---> 2 H2O + 2 CO2 H = entalpies of the products x coefficien - enthalpies of the reactant x coefficients [2 x (-393.5) + 2 x (-285.9)]-[1 x -unknown + 3 x 0]= -1411.1 sove for C2H4 = 52.9 kJ/mol
Answered by Eternal Flame - Sat Nov 8 17:18:53 2008
Q. The standard enthalpy of combustion of ethene gas, C2H4(g), is -1411.1 kJ/mol at 298 K. Given the following enthalpies of formation, calculate Hf for C2H4(g). CO2(g) -393.5 kJ/mol H2O(l)-285.9 kJ/mol
Asked by NoName - Sat Nov 8 15:39:20 2008 - - 1 Answers - 0 Comments
A. Firs write the balanced equation: C2H4 + 3 O2 ---> 2 H2O + 2 CO2 H = entalpies of the products x coefficien - enthalpies of the reactant x coefficients [2 x (-393.5) + 2 x (-285.9)]-[1 x -unknown + 3 x 0]= -1411.1 sove for C2H4 = 52.9 kJ/mol
Answered by Eternal Flame - Sat Nov 8 17:18:53 2008
what is the standard enthalpy change of combustion of ethanol and iso-butanol?
Q. I can't seem to find it anywhere :[
Asked by -a- - Sun Sep 21 17:33:43 2008 - - 1 Answers - 0 Comments
Q. I can't seem to find it anywhere :[
Asked by -a- - Sun Sep 21 17:33:43 2008 - - 1 Answers - 0 Comments
What is the standard enthalpy of formation of NH4Cl (aq)?
Q. I can't find it anywhere I can only find it as a solid and I desperately need to find it. Thanks so much
Asked by dramafreak:) - Mon Oct 27 10:34:51 2008 - - 1 Answers - 1 Comments
A. -300. i think you can just look up the standard enthalpy for the two separate ions and add them up. NH4 (aq) is -132.8 and Cl (aq) is 167.2...so when you add them up you get -300.0. i hope this helps.
Answered by anonymous - Wed Oct 29 13:48:00 2008
Q. I can't find it anywhere I can only find it as a solid and I desperately need to find it. Thanks so much
Asked by dramafreak:) - Mon Oct 27 10:34:51 2008 - - 1 Answers - 1 Comments
A. -300. i think you can just look up the standard enthalpy for the two separate ions and add them up. NH4 (aq) is -132.8 and Cl (aq) is 167.2...so when you add them up you get -300.0. i hope this helps.
Answered by anonymous - Wed Oct 29 13:48:00 2008
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