chem Jan 27 delta H and heat transfer calculations
1
Jan 2111:35 PM
NaNO3 Na+ + NO3
When a 15 gram sample of solid sodium nitrate (NaNO3) dissolves in 125 grams of water in a coffee cup calorimeter, the temperature drops from 25 C to 16.4 C. Calculate the delta H(in KJ/mol for sodium nitrate) c for water = 4.19 J/gC
Jan 2111:38 PM
chem Jan 27 delta H and heat transfer calculations
2
Jan 2111:38 PM
9.0 grams of charcoal (Carbon) were completely consumed in a bomb calorimeter. If we assume that the 2.0 L of water absorbed all of the heat released by the charcoal, and if the temperature of the water increased from 20.25 to 56.04C, what is the molar enthalpy of carbon? c for water = 4.19 J/[g C]
Jan 2111:39 PM
Qwater = mc T= (2000g)(4.19 J/gC)(56.0420.25)Qwater = 299 920 JQreaction = 299 920 J
( H) = (Qreaction)(Molar mass of reactant) / (mass of reactant) = ( 299 920)(12.011 g/mole )/ (9.0g) = 400 260.17 J/mol of Carbon
chem Jan 27 delta H and heat transfer calculations
3
Jan 2111:43 PM
450 mL of water calorimetre contain the following neutralization reaction: AB + DC AC + DCWe go from an initial temperature of 22.3 oC to a maximum of 29.2 C. Calculate the molar heat(enthalpy) of neutralization if the mass of the reactants is 0.2g and the molar mass is 16g/mol.
c for water = 4.19 J/[g C]
Jan 2111:46 PM
Qwater = mc T= (450 g)(4.19)(29.222.3)Qwater =13009.95 JQreaction = 13009.95 J
Hreaction = (Qreaction)(Molar mass of reactant)/ (mass of reactant) ( 13009.95 J) (16g/mol)/0.2g = 1 040 796 J/mol
chem Jan 27 delta H and heat transfer calculations
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Jan 2111:54 PM
Find the final temperature of the following mixture:400. g of Cu initially at 99.0 oC25 L of water initially at 10.0 oCc for Cu = 0.39 J/[g C]c for water = 4.19 J/[g C]
Jan 279:24 PM
chem Jan 27 delta H and heat transfer calculations
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Jan 279:18 PM
A 323 g sample of brass at 112C is placed ina calorimeter cup that contains 265 g of waterat 34C.If you disregard the absorption of heatby the cup, what will be the final temperature? Assume the specific heat of brass is0.376 J/g.C, and the specific heat of wateris 4.19 J/g·C.
Jan 2111:54 PM
Heat Transfer:
Q = mcΔT
Qbrass = Qwater
((m)(c)(ΔT)) = (m)(c)(ΔT)
((323g) (0.376J/kg.C) (Tf 112C)) = (265) (4.19J/g.C) (Tf 34)
((121.448) (Tf 112C)) = (1110.35) (Tf 34)
(121.448Tf 13602.176) = 1110.35Tf 37751.9
121.448Tf + 13602.176 = 1110.35Tf 37751.9
13602.176 + 37751.9 = 1110.35Tf + 121.448Tf
51354.076= 1231.798T f
51354.076/1231.798 = T f
Tf = 41.69 ºC
chem Jan 27 delta H and heat transfer calculations
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Jan 257:12 PM
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