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enthalpy change calculator
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From Wikipedia
Enthalpy change of solution  Wikipedia, the free encyclopedia
The enthalpy of solution, enthalpy of dissolution, or heat of solution is the ... potassium chlorate, +41.38. acetic acid, 1.51. sodium hydroxide, 44.51 ...
From Yahoo Answers
Question:use the enthalpy changes for the composoiton of aluminum and iron
2AL(s)+3/2O2(g) Al2O3(s)
H=1669.8KJ
2Fe(s) + 3/2 O2(g) Fe2O3
H= 824.2KJ
COULD YOU PLEASE SHOW WORKING ON HOW YOU GOT YOUR ANSWER?
Answers:I think you have made a mistake in typing the equation. The equation should be: 2Al + Fe2O3 > Al2O3 + 2Fe {The aluminothermic process for extraction of iron} According to Hess' law, the chemical equations can be added or subtracted as if they were algebraic equations. We need to express the above equation in the form of the 2 equations the data of which is provided. Reversing the second equation, we have Fe2O3 > 2Fe + 3/2 O2 { H = +824.2 kJ} Adding this to the first equation, we have 2Al + Fe2O3 +3/2 O2 > 2Fe + 3/2 O2 + Al2O3 Cancelling the commom O2, 2Al + Fe2O3 > 2Fe + Al2O3. H(reaction) = 1669.8+824.2 =845.6 kJ
Answers:I think you have made a mistake in typing the equation. The equation should be: 2Al + Fe2O3 > Al2O3 + 2Fe {The aluminothermic process for extraction of iron} According to Hess' law, the chemical equations can be added or subtracted as if they were algebraic equations. We need to express the above equation in the form of the 2 equations the data of which is provided. Reversing the second equation, we have Fe2O3 > 2Fe + 3/2 O2 { H = +824.2 kJ} Adding this to the first equation, we have 2Al + Fe2O3 +3/2 O2 > 2Fe + 3/2 O2 + Al2O3 Cancelling the commom O2, 2Al + Fe2O3 > 2Fe + Al2O3. H(reaction) = 1669.8+824.2 =845.6 kJ
Question:The activation energy for a reaction in the forward direction is 78 kJ. THe activation energy for the same reaction in reverse is 300 kJ. If the energy of the products is 25 kJ, then:
A) What is the energy of the reactants?
B) What is the enthalpy change for the forward reaction?

it would be appreciated if you could also explain the approach to the problems.
THANKS
Answers:Draw a potential energy diagram. Something like this.http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif Now I assume when they say "the energy of the products is 25 KJ" they're talking about the forward reaction. So if the products are at 25 KJ and the activation energy of the reverse reaction is 300 KJ then the peak is at 300+25=325 KJ. Now if the activation energy of the forward reaction is 78 KJ then that means it takes 78 KJ for the forward reaction to get to the peak at 325 KJ so then the forward reaction must have started at 32578=247 KJ. That answers question A. A) 247 KJ B) H=energy of products minus energy of reactants =25 KJ247 KJ=222 KJ Here's a badly drawn potential energy diagram to help you understand. http://s171.photobucket.com/albums/u293/jfdnew/?action=view¤t=potentialenergydiagram.jpg There you go. :)
Answers:Draw a potential energy diagram. Something like this.http://www.bbc.co.uk/scotland/education/bitesize/higher/img/chemistry/calculations_1/pe_diags/fig10.gif Now I assume when they say "the energy of the products is 25 KJ" they're talking about the forward reaction. So if the products are at 25 KJ and the activation energy of the reverse reaction is 300 KJ then the peak is at 300+25=325 KJ. Now if the activation energy of the forward reaction is 78 KJ then that means it takes 78 KJ for the forward reaction to get to the peak at 325 KJ so then the forward reaction must have started at 32578=247 KJ. That answers question A. A) 247 KJ B) H=energy of products minus energy of reactants =25 KJ247 KJ=222 KJ Here's a badly drawn potential energy diagram to help you understand. http://s171.photobucket.com/albums/u293/jfdnew/?action=view¤t=potentialenergydiagram.jpg There you go. :)
Question:You are provided with an unknown acid whose relative forumla mass is 135. The acid is a white crystaline solid, and is very soluble in water. It is also monobasic. The acid will be neutralised with sodium hydroxide. I dont know what calculations to do once the enthalpy change is found.
thanks for the healp
Answers:Enthalpy change calculation is q = mc (delta) T where q is the enthalpy change, m is the mass of the solution, c is the heat capacity and (delta) T is the change is temperature extrapolated from the graph. If m = 100, c = 4.18 and you get a temp. change of, say, 6 degrees, then q = 100 x 4.18 x 6 = 2508J Firstly, you need to know the number of moles of solute (the acid) you used. If you weighed out 25g of the unknown acid to be dissolved, then you do Mass / Mr = 25 / 135 = 0.185M of solute. Chances are is that you dissolved this into 250cm3 of water, then took 50cm3 of that solution and added it to 50cm3 of alkali. If this is the case, then you added one fifth of your made up acid solution to the alkali (as 50 is one fifth of 250) so you added one fifth of the dissolved acid. This means you effectively added 0.185 / 5 = 0.037M of the solute to alkali and 50cm3 of water. With this knowledge, you can now calculate the molar enthalpy change. You know how much enthalpy change 0.037M of the acid produces, so you just work out how many 0.037s you can fit in 1.00, which is 1 / 0.037 = 27.07. Soooo, finally, to get your Molar Enthalpy Change, you multiply your 0.037 Enthalpy Change by 27.07. Convert this the kJmol1 by dividing by 1000. But remember, the enthalpy change should be a NEGATIVE number because the solution is LOSING heat to the enviroment, so make sure to explain this in your c/w and put the negative number in.
Answers:Enthalpy change calculation is q = mc (delta) T where q is the enthalpy change, m is the mass of the solution, c is the heat capacity and (delta) T is the change is temperature extrapolated from the graph. If m = 100, c = 4.18 and you get a temp. change of, say, 6 degrees, then q = 100 x 4.18 x 6 = 2508J Firstly, you need to know the number of moles of solute (the acid) you used. If you weighed out 25g of the unknown acid to be dissolved, then you do Mass / Mr = 25 / 135 = 0.185M of solute. Chances are is that you dissolved this into 250cm3 of water, then took 50cm3 of that solution and added it to 50cm3 of alkali. If this is the case, then you added one fifth of your made up acid solution to the alkali (as 50 is one fifth of 250) so you added one fifth of the dissolved acid. This means you effectively added 0.185 / 5 = 0.037M of the solute to alkali and 50cm3 of water. With this knowledge, you can now calculate the molar enthalpy change. You know how much enthalpy change 0.037M of the acid produces, so you just work out how many 0.037s you can fit in 1.00, which is 1 / 0.037 = 27.07. Soooo, finally, to get your Molar Enthalpy Change, you multiply your 0.037 Enthalpy Change by 27.07. Convert this the kJmol1 by dividing by 1000. But remember, the enthalpy change should be a NEGATIVE number because the solution is LOSING heat to the enviroment, so make sure to explain this in your c/w and put the negative number in.
Question:the enthalpy of fusion of chlorine is 6.4 and the entropy of fusion is 37.2
Answers:Set delta G = 0, and T = delta H/delta S Try 6,400/37.2. Don't forget that the enthalpy change was in kJ/mol.
Answers:Set delta G = 0, and T = delta H/delta S Try 6,400/37.2. Don't forget that the enthalpy change was in kJ/mol.
From Youtube
Enthalpy Changes for Chemical Reactions :General Chemistry lecture covering endothermic and exothermic reactions, Hess Law, bond energies, and standard enthalpies of formation for chemical substances. We also describe the method of calculating the enthalpy of combustion for materials.
Enthalpy of Expansion and Compression of Gases :General Chemistry lecture covering the basic transfer of heat and work to and from gases during expansion and compression. We discuss how to compute heat and/or work for isochoric heating, isothermal expansion and adiabatic expansion. We also present a general strategy for determining changes in thermodynamic state functions for arbitrary changes in pressure, volume and temperature for ideal gases.