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# Chemistry Help - Enthalpy

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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 20.6Â°C to 53.6Â°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 by wapiti
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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) â†’ 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 Ã· 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

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thank you so much!!!!!! (: (: (:

:laugh:  :laugh:  :laugh:  :coolgleamA:  :thumbup1:

(That's thanks from dd.  and thanks from me too!)

Edited by wapiti
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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) â†’ 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 Ã· 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

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!

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Thank you, Dicentra!  Dd will appreciate the explanation!

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