Part A A calorimeter contains 35.0 mL of water at 12.0 C . When 1.60 g of X (a s

Question

Part A

A calorimeter contains 35.0 mL of water at 12.0 C . When 1.60 g of X (a substance with a molar mass of 55.0 g/mol ) is added, it dissolves via the reaction

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

and the temperature of the solution increases to 25.0 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/(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.

Part B

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

Mass of X = 1.60g
moles of X = m / molar mass = 1.60/55 = 0.03 moles
Mass of solution (m ) = 35 g
Intital temperature T1 = 12.0 °C
Final temperature T2 = 25.0 °C
Change in temperature T = T2- T1
=25.0 - 12.0
=13.0 °C
Specific heat capacity of water Cp =4.18 J/g°C
Heat generated
Q = m x Cp x T ……….. (1)
Substitute values in this formula we get
Q = 35 x 4.18 x 13.0 J

Q= 1901.9 J
Divide by 1000 to convert in Kj we get
Q= 1.902 kJ
Change in enthalpy
H = Q / number of moles
H = 1.902 kj/ 0.03 mol
H = 63.4 Kj/ mol
This reaction is producing energy so it a exothermic reaction
And in exothermic reaction H is always negative
So H = - 63.4 Kj/ mol

Part B

The heat produced in a bomb calorimeter when combusting a given amount of known substance is the internal energy of the substance combusted per the number of moles of the substance that was burned.

10.0 g of sucrose = (10.0 g) / (342.3 g/mol) = 0.02921 moles of sucrose

Heat capacity of the calorimeter = 7.50 kJ/ deg C.

This means that the temperature of calorimeter increases 1 deg C when 7.50 kJ of heat is absorbed by the calorimeter from the combustion process.

Since the temperature increase was 22.0 deg C when combusting the 0.02921 moles of sucrose, then the combustion process must have given off (22.0 deg C)(7.50 kJ/deg C) = 165.0 kJ..

Therefore, the change in internal energy per mole of sucrose combusted would be = (165.0 kJ) / (0.02921)moles of sucrose combusted) = 5648.7 kJ/mol of sucrose combusted.

5648.7 kJ/mol = 5.65 ×10^3

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