wapiti Posted March 13, 2017 Share Posted March 13, 2017 (edited) This is wapiti's dd. Can anyone help me with these chemistry enthalpy questions? i tried for so long and i can't figure them out. thank you so much!!! 1.) A 7.06 g sample of methane, CH4, is burned in excess oxygen producing carbon dioxide gas and liquid water. The reaction occurs in a calorimeter surrounded by 1.52 kg of water resulting in a temperature change from 20.6°C to 53.6°C. What is the heat of combustion of methane? Include the sign in your answer. Assume the heat released by the reaction is completely absorbed by the water. 2.) 13.5 g of an unknown metal at 91.4 oC is placed in 79.0 g of water (s = 4.18 J/g-oC) at 11.1 oC. What is the specific heat of the metal if thermal equilibrium is reached at 11.8 oC? Edited March 13, 2017 by wapiti Quote Link to comment Share on other sites More sharing options...
Arcadia Posted March 13, 2017 Share Posted March 13, 2017 Assuming a closed system for (2) Heat absorbed by water = 4.18J/g degC * (11.8-11.1)degC * 79.0g = 231.154 Joules Specific heat of metal = 231.154J / (91.4-11.8) /13.5g = 0.215 J/g degC For (1), my answer is off from the known heat of combustion of methane 890 kJ/mole. I see if I can figure out where I went wrong later. Or Dicentra can help :) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Calculate moles of methane used molar mass of methane = 16.043 g/mol mass of methane used = 7.06g Moles of methane used = 0.44 moles Calculate energy required to change temperature of water in calorimeter energy = specific heat capacity of water x mass of water x change in water temperature energy = 4.186 J K-1g-1 x 1520 g x 33 degC = 209.970 kJ Assume all the heat produced from burning ethanol has gone into heating the water, 0.44 mole of methane produced 209.970 kJ of heat. 1 mole of methane would produce 209.970 kJ ÷ 0.44 mol = 477.204 kJ mol-1 For (1), I used these webpages as I was too lazy to go get my chem book Simple http://www.ausetute.com.au/heatcomb.html More details http://cbc-wb01x.chemistry.ohio-state.edu/~woodward/ch121/ch5_enthalpy.htm Quote Link to comment Share on other sites More sharing options...
wapiti Posted March 13, 2017 Author Share Posted March 13, 2017 (edited) thank you so much!!!!!! (: (: (: :laugh: :laugh: :laugh: :coolgleamA: :thumbup1: (That's thanks from dd. and thanks from me too!) Edited March 13, 2017 by wapiti 1 Quote Link to comment Share on other sites More sharing options...
Dicentra Posted March 14, 2017 Share Posted March 14, 2017 Assuming a closed system for (2) Heat absorbed by water = 4.18J/g degC * (11.8-11.1)degC * 79.0g = 231.154 Joules Specific heat of metal = 231.154J / (91.4-11.8) /13.5g = 0.215 J/g degC For (1), my answer is off from the known heat of combustion of methane 890 kJ/mole. I see if I can figure out where I went wrong later. Or Dicentra can help :) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) Calculate moles of methane used molar mass of methane = 16.043 g/mol mass of methane used = 7.06g Moles of methane used = 0.44 moles Calculate energy required to change temperature of water in calorimeter energy = specific heat capacity of water x mass of water x change in water temperature energy = 4.186 J K-1g-1 x 1520 g x 33 degC = 209.970 kJ Assume all the heat produced from burning ethanol has gone into heating the water, 0.44 mole of methane produced 209.970 kJ of heat. 1 mole of methane would produce 209.970 kJ ÷ 0.44 mol = 477.204 kJ mol-1 For (1), I used these webpages as I was too lazy to go get my chem book Simple http://www.ausetute.com.au/heatcomb.html More details http://cbc-wb01x.chemistry.ohio-state.edu/~woodward/ch121/ch5_enthalpy.htm Well done, Arcadia! :) The only thing I would change is the treatment of the signs in your calculations. Oh - and I get the same value as you do for #1. I don't think we're wrong in the sense that we simply used the numbers provided in the question. :) The fact that it doesn't match the tabled value for the enthalpy of combustion of methane is probably just due to wonky numbers being given in the question. :) When heat is being transferred from one substance to another, the convention in thermodynamics is to have the heat lost be given a negative sign and the heat absorbed be given a positive sign. This makes sense and can be shown to be true if one always remembers that delta T = Tfinal - Tinitial. For #1, then, when we calculate the energy absorbed by the water, the sign would be positive because Q=(1520g)(4.18J/goC)(53.6oC-20.6oC)=+2.10x105J. This means that the methane lost an equivalent amount of energy so Q = -2.10x105J per 0.441mol of methane. Using the same math as Arcadia, the heat of combustion per mole of methane would then be -4.76x105J/mol - don't forget the negative! For #2, the energy absorbed by the water would be Q=(79.4g)(4.18J/goC)(11.8oC-11.1oC)=+231J. This means that, for the metal, Q = -231J. When we rearrange the specific heat equation to solve for C (specific heat), we would have C = -231J/(13.5g x -79.6oC). The reason the change in temp is negative is because delta T = Tfinal - Tinitial so delta T = 11.8-91.4 = -79.6oC. The signs, though, all work out when we finish the calculation for C. From above, C = -231J/(13.5g x -79.6oC) = 0.215J/goC. This gives a positive value for C (which it should be). Hope that helps! 2 Quote Link to comment Share on other sites More sharing options...
wapiti Posted March 14, 2017 Author Share Posted March 14, 2017 Thank you, Dicentra! Dd will appreciate the explanation! 1 Quote Link to comment Share on other sites More sharing options...
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