Triangle H is a state function. Calculate heat of combustion of methane by the p

Question

Triangle H is a state function. Calculate heat of combustion of methane by the pathways below CH_4 + 2O_2(g) a CO_2(g) = 2H_2O(g) Triangle H degree = ? Direct measurement by isobaric calorimetry: triangle H_rxn = q_rxn = (q_H2O + q_cal) 3.785 g of methane, CH_4, is burned in excess oxygen in a calorimeter containing 2.50 kg of water resulting in a temperature change from 24.7degree C to 42.8 degree C. what is the heat of combustion of methane (kJ/mol.)? (assume no heat loss to the calorimeter i.e q_cat = 0; specific heat capacity of water is 4.18J/gdegree C) Using the triangle H's of the following related reactions: (hess Law) 2O(g) a O_2(g) triangle Hdegree = -249 kj/mol 2H(g) + O(g) à H_2O(g) triangle H degree = -800kJ/mol C(graphite) + 2O(g) à CO_2(g) triangle H degree = -643 kJ/mol C(graphic) + 2H_2(g) à CH_4(g)triangle H degree = -75 kJ/mol 2H(g) à H_2(g) triangle H degree = -434kJ/mol Using standard heats of formation: triangle H degree = sigma(ntriangleH_fdegree) products - sigma(n triangle H_f degree) reactants

Explanation / Answer

The energy releases by reaction (i.e., heat of combustion) is absorbed by water, so it can be found from the temperature rise of the water:
Energy = mC(delta T) =
E = (2.50kg) x (4.184 kJ/kg/degree) x (42.8-24.7) =
E= -189.32 kJ (negative, hence it is exothermic)

The amount of Methane that reacted is (3.785g)/(16.0g/mol) = 0.2365 mol.
The heat of combustion is then -189.32kJ/0.2365mol =
-800.50 kJ/mol.

2) The combustion of CH4 is given by, CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(g)

Enthalpy change, dH is given by, dH0 = (ndH0products)-(ndH0Reactants)

dH0 CH4 = (-643+(2*-800))-(-75+2*0) = -2243-(-75) = -2168KJ/mol

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