A.) A total of 2.00 mol of a compound is allowed to react with water in a foam c

Question

A.) A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 100 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water.

B.)A calorimeter contains 20.0 mL of water at 15.0 C . When 1.80 g of X (a substance with a molar mass of 67.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)X(aq)

and the temperature of the solution increases to 26.5 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/(gC)], 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.

C.)

Consider the reaction

C12H22O11(s)+12O2(g)12CO2(g)+11H2O(l)

in which 10.0 g of sucrose, C12H22O11, 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

A) (4.184 J/g·°C) x (100 g) x (24.70 - 21.00)C = 1548.08 J / 2 mol = 774.04 J (or) 0.774 kJ

B) 1. (4.18 J/gC) *( 20.0 g H2O + 1.80g X)* (26.5 C-15 C) = 1047.926 J
2. 1047 / (1.80g/67.0g/mole) = - 8.689

C) heat generated = 22.0 C * 7.50 kJ/C = 165 kJ

kJ/mole sucrose = 165 kJ * (10.0 g sucrose / 342.29 g/mole) = 4.82 kJ /mole

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