Part A A calorimeter contains 25.0 mL of water at 11.5 C . When 1.20 g of X (a s

Question

Part A A calorimeter contains 25.0 mL of water at 11.5 C . When 1.20 g of X (a substance with a molar mass of 73.0 g/mol ) is added, it dissolves via the reaction X(s)+ H 2 O(l)X(aq) and the temperature of the solution increases to 30.0 C . Calculate the enthalpy change, H , for this reaction per mole of X . Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g C) ], that density of water is 1.00 g/mL , and that no heat is lost to the calorimeter itself, nor to the surroundings. Express the change in enthalpy in kilojoules per mole to three significant figures.

Part B Consider the reaction C 12 H 22 O 11 (s)+12 O 2 (g)12C O 2 (g)+11 H 2 O(l) in which 10.0 g of sucrose, C 12 H 22 O 11 , 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 energy, E , for this reaction per mole of sucrose. Express the change in internal energy in kilojoules per mole to three significant figures.

Explanation / Answer

Mass of the total solution = mass of water + mass of substance = 25ml*1g/ml+1.2= 26.2 gm

From the reaction, enthalpy of neutralization = mass* specific heat* temperature difference= 26.2*4.18*(30-11.5) =2026 joules

This is dues to addition of 1.2 gm of X of molar mass = 73 g/mol, moles of X= 1.2/73=0.0164

Heat of neutralization = 2026/0.0164= 123248.3 Joules

2.

Enthalpy change = specific heat of calorimeter* temperature increase= 7.5*22 KJ=165 KJ

Moles of sucrose = mass/molar mass, molar mass of sucrose(C12H22O11)= 12*12+22+11*16= 342

Moles of sucrose = 10/342= 0.029, enthalpy change= 165/0.029 =5690 Kj/mole

From enthalpy change, deltaH= deltaU+deltanR*T , deltaU= change in internal energy, deltan= change in no of moles of gas = 12-12= 0

5690= deltaU + 0

delltaU= 5690 Kj/mole

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.