Question

A calorimeter contains 20.0 mL of water at 12.5 ∘C . When 1.40 g of X (a substance with a molar mass of 64.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)→X(aq)

and the temperature of the solution increases to 28.5 ∘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.

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.

Answer 1

1)

weare given the following values
Mass of reactant= 1.40g
moles of reactant = m / molar mass = 1.40 / 64 = 0.0218 moles
Mass of solution (m ) = 20g
Intital temperature T1 = 12.5 °C
Final temperature T2 = 28.5°C
Change in temperature ∆T = T2- T1
=28.5 - 12.5 =
=16 °C
Specific heat capacity of water Cp =4.184 J/g°C
Heat generated
Q = m x Cp x ∆T ……….. (1)
Plug the values in this formula we get


Q = 20 x 4.184 x 16 j
Q= 1338.88 j

Divide by 1000 to convert in Kj we get

Q= 1.339 kj
Change in enthalpy
∆H = Q / number of moles
∆H = 1.339 kj/ 0.0218 mol
∆H = 61.4 Kj/ mol
This reaction is producing energy so it a exothermic reaction
And in exothermic reaction ∆H is always negative
So our answer will ∆H = -61.4 Kj/ mol

2)

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.

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

You are given that the heat capacity of the calorimeter is 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) = 5649 kJ/mol of sucrose combusted.