Question

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.9322.93 °C to 27.4327.43 °C. What is the internal energy change, Δ????ΔU , for the combustion of 1.5011.501 g of fructose?

Answer 1

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0   (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

Answer 2

The combustion of 1.501 g of fructose, C₆H₁₂O₆(s), in a bomb calorimeter with a heat capacity of 5.20 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.93 °C to 27.43 °C. What is the internal energy change, ΔU, for the combustion of 1.501 of fructose?

The combustion of 1.501 g of fructose, C₆H₁₂O₆(s), in a bomb calorimeter with a heat capacity of 5.20 kJ/°C, resulting in an increase in the temperature of the calorimeter and its contents from 22.93 °C to 27.43 °C, represents a change in the internal change for the combustion of -2.81 × 10³ kJ/mol.

A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction and in determining the internal energy change (ΔU) of the combustion process.

First, we will calculate the heat absorbed by the bomb calorimeter (Qcal) using the following expression.

$Qcal = C \times \Delta T$

where,

• C: heat capacity of the calorimeter

• ΔT: change in the temperature

$Qcal = \frac{5.20 kJ}{\° C } \times (27.43 \° C - 22.93 \° C) = 23.4 kJ$

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter and the heat released by the combustion (Qcomb) is zero.

$Qcal + Qcomb = 0\\Qcomb = -Qcal = -23.4 kJ$

1.501 g of fructose with a molar mass of 180.16 g/mol reacted. The corresponding number of moles are:

$n = 1.501 g \times \frac{1mol}{180.16g} = 8.331 \times 10^{-3} mol$

Finally, we can calculate the change in the internal energy for the combustion reaction of fructose using the following expression.

$\Delta U = \frac{Qcomb}{n} = \frac{-23.4 kJ}{8.331 \times 10^{-3} mol} = -2.81 \times 10^{3} kJ/mol$

The combustion of 1.501 g of fructose, C₆H₁₂O₆(s), in a bomb calorimeter with a heat capacity of 5.20 kJ/°C, resulting in an increase in the temperature of the calorimeter and its contents from 22.93 °C to 27.43 °C, represents a change in the internal change for the combustion of -2.81 × 10³ kJ/mol.

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