Question

<P 8 Calorimetry < 7 of 29 > A calonmeter is an insulated device in which a chemical reacton is contained. By measuring the temperature change, A7, we can calculate the heat released or absorbed during the reaction using the following equation: Review Constants Periodic Table A calorimeter contains 18.0 mL of water at 11.5 °C. When 2.30 g of X (a substance with a molar mass of 46.0 g/mol) is added, it dissolves via the reaction X(s) + II, O(1) X(aq) and the temperature of the solution increases to 28.5 "C. Calculate the enthalpy change. AH, for this reaction per mole of X mL and that no heat is lost to the 9-specific heat x mass X AT Or, if the calorimeter has a predetermined heal capacity, C, the equation becomes q=C XAT At constant pressure, the enthalpy change for the reaction, Alis equal to the heat, that is Assume that the specific heat of the resulting solution is equal to that of water 4.18 J/(.'C), that density of water is 1.00 calorimeter itsell, nor to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures. View Available Hint(s) AN-4 IVO AXCP R O ? but it is usually expressed per mole of reactant and with a sign opposite to that of g for the surroundings. The total internal energy change, AE sometimes referred to as AU), is the sum of heat, q. and work done, w AII = k.l/mol AE 9w = 0 Submit However, at constant volume (as with a bomb calorimeter) and so AE=Q Part B Consider the reaction C2H2011(s) | 120, (g) >1200, (g) | 11H2O(1) in which 10.0 F of sucrose, C2H011, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/°C. The temperature increase inside the calorimeter was found to be 22.0°C.Calculate the change in internal cricrgy, AE, for this reaction per mole of sucrose. Express the change in internal energy in kilojoules per mole to three significant figures. View Available Hinta) AXO O ? AE- kJ/mol Submit

Answer 1

Heat gained by system to raise it's temperature = m.s.∆T = (18.0+2.30)g×4.18J/g.℃×(28.5-11.5)℃ = 1442.518J

Number of moles of X dissolved = 2.30g/46.0g/mol= 0.0500mol

∆H = - 1442.518J/0.050mol = -28850.36J/mol = - 28.85 kJ/mol

= - 28.9 kJ/mol. (Answer)

Part B:

Number of moles of sucrose = 10.0g/342.2965g/mol = 0.02921mol

Heat released = 7.50kJ/℃ × 22.0℃ = 165kJ

∆E = - 165kJ/0.02921mol = - 5647.9 kJ/mol

∆E = - 5650 kJ/mol