When 1.04g of cyclopropane was burnt in excess oxygen in a bomb calorimeter, the temperature rose by 3.69K. The total heat capacity of the calorimeter and it's contents was 14.01kJ/K. Determine the enthalpy of combustion of cyclopropane.

In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:

[tex]Q_{rxn}+Q_{cal}=0[/tex]

Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:

[tex]Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ[/tex]

Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:

[tex]n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}[/tex]

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mahima6824soni mahima6824soni · Answer 2 · 2022-04-07T15:05:47+02:00

The enthalpy of combustion of cyclopropane is [tex]\Delta _c_o_m_b H= - 2.093 kJ/mol[/tex]

What is the enthalpy of combustion?

Enthalpy of combustion is the amount of heat produced when one mole of substance completely burns.

By the formula of calorimetry

[tex]Q_r_x_n + Q_c_a_l = 0[/tex]

The temperature change and total heat capacity is

[tex]Q_r_x_n = -14.01 KJ/K \times 3.69 K\\\\Q_r_x_n = -51.70 KJ[/tex]

Now, we get the enthalpy of combustion by dividing by the moles of 1.04 g of cyclopropane.

The number of moles is 42.09 g/mol.

[tex]n = \dfrac{1.4}{42.09} = 0.0247\\\\\Delta_c_o_m_bH =\dfrac{Q_r_x_n}{n} \\\\\Delta_c_o_m_bH = -2.093\; kJ/mol[/tex]

Thus, the enthalpy of combustion is -2.093 kJ/mol

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