Question

Part 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 101 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. Enter your answer in kilojoules per mole of compound to three significant figures.

Part B)

A calorimeter contains 24.0 mL of water at 13.0 ∘C . When 1.60 g of X (a substance with a molar mass of 59.0 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, ΔH, 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 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, ΔE, for this reaction per mole of sucrose.

Express the change in internal energy in kilojoules per mole to three significant figures.

Part D)

What is the enthalpy for reaction 1 reversed?

reaction 1 reversed: CO2→CO + 12O2

Express your answer numerically in kilojoules per mole.

Answer 1

1) using formulae

Q= m * C * T

= 101 g * 4.18 J/g⁰C * (24.7⁰C - 21⁰C)

= 1562.066 J = 1.56 KJ

2)   

  Q= m * C * T mass of water = 24 g ( as density for water is equal to 1)

= 24g * 4.18 J/g⁰C * 26.5⁰C -13⁰C

= 1354.32 J = 1.35 KJ /mole

3) we calculate the number of moles of sucrose = given mass / molar mass

= 10 g /(342.3 g/mol) =  0.02921 moles of sucrose

As the heat capacity of the calorimeter is 7.50 kJ/ deg C .

(22.0 deg C)(7.50 kJ/deg C) = 165.0 kJ.. ( the temperature increase was 22.0 deg C when combusting the 0.02921 moles of sucrose )

the change in internal energy = (165.0 kJ) / (0.02921 moles of sucrose combusted) = 5648.75 kJ/mol=

564*10 kJ/mol