CH%20301%20Chapter%206%20calorim%20example%20v2

# CH%20301%20Chapter%206%20calorim%20example%20v2 - 0.01408...

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Data: Using a bomb calorimeter of heat capacity 25.2 J/ o C, containing 500 ml of water, 1.10g of benzene was burned. The temperature of the water went from 23.00 o C to 44.72 o C. Use this data to find H combustion for benzene. The specific heat of water = 4.184 J/g 0 C. 1. how much heat was absorbed by water: q = 500ml * 1g/ml * 4.184 J/ g 0 C * (44.72-23.00 o C) = 45438.24 J 2. how much heat was absorbed by calorimeter: q = 25.2 J/ o C * (44.72-23.00 o C) = 547.344 J Total: q (calorimeter + water) = 45438.24 J + 547.344 J = 45985.584 J = 45.985584 k J = - q reaction q reaction = - 45.985584 kJ As this is a bomb calorimeter we have found U. We need H. Find out how much C 6 H 6 was burned: 1.10g * 1mol / 78.12 g = 0.01408 mol of C. Example: Getting Thermochemical Equation (Reaction Enthalpy) from Bomb Calorimetry Data
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Unformatted text preview: 0.01408 mol of benzene when burned has ∆ U of - 45.985584 kJ To convert between ∆ U and ∆ H and quote ∆ H combustion we need the standard combustion reaction for benzene: C 6 H 6(s) + 15/2 O 2(g)--> 6CO 2(g) + 3H 2 O (l) 1 mole of benzene is burned per 1 mole of reaction Find ∆ U for 1 mole of reaction: - 45.985584 kJ / 0.01408 mol benzene * (1mol benzene / 1 mol reaction) = - 3266kJ/ 1mol reaction Now convert between ∆ U and ∆ H: See above: ∆ n = (6 -15/2) = -1.5 ∆ H combustion = ∆ U combustion + ∆ nRT = - 3266kJ + (-1.5mol * 8.314 J/mol.K * 298K) ∆ H combustion = - 3266kJ - 3.716kJ = - 3270kJ/mol Example: Getting Thermochemical Equation (Reaction Enthalpy) from Bomb Calorimetry Data...
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