Answer:
[tex]\Delta _{comb}H=-2,093\frac{kJ}{mol}[/tex]
Explanation:
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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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The enthalpy of combustion of cyclopropane is [tex]\Delta _c_o_m_b H= - 2.093 kJ/mol[/tex]
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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