Question

The standard enthalpy of combustion of cyclopropane is -2091 kJ/mole at ${{25}^{o}}C$ then calculate the enthalpy of formation of cyclopropane. If $\Delta H_{f}^{o}(C{{O}_{2}})$ = -393.5 kJ/mol and $\Delta H_{f}^{o}({{H}_{2}}O)$ = -285.8 kJ/mol.

Answer 1

If it is necessary to add two chemical equations and come up with another equation according to Hess’s law, the enthalpy of the third equation is the sum of the first two equations. Based on the consequences of the first law of thermodynamics, enthalpy is a state function.

Complete step by step solution:
The reaction of formation of cyclopropane,
$3C(graphite)+3{{H}_{2}}(g)\to {{C}_{3}}{{H}_{6}}(cyclopropane)$ --- (1)
The combustion reaction of cyclopropane,
${{C}_{3}}{{H}_{6}}(g)+\dfrac{9}{2}{{O}_{2}}\to 3C{{O}_{2}}(g)+3{{H}_{2}}O(g)$ ---- (2)
Standard enthalpy of combustion of cyclopropane, $\Delta H_{c}^{o}$ = -2091 kJ/mole at ${{25}^{o}}C$
Another way of formation cyclopropane is the reverse of the combustion reaction,
$3C{{O}_{2}}(g)+3{{H}_{2}}O(g)\to {{C}_{3}}{{H}_{6}}(g)+\dfrac{9}{2}{{O}_{2}}(g)$ --- (3)
Given that standard enthalpy of formation of $C{{O}_{2}}\And {{H}_{2}}O$ ,
$C(graphite)+{{O}_{2}}(g)\to C{{O}_{2}}(g),\Delta H_{f}^{o}(C{{O}_{2}})=-393.5kJ/mol$--- (5)
${{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(l);\Delta H_{f}^{o}({{H}_{2}}O)=-285.8kJ/mol$ --- (6)
According to Hess’s law,
Enthalpy of formation of cyclopropane = $3\times \Delta H_{f}^{o}(C{{O}_{2}})+3\times \Delta H_{f}^{o}({{H}_{2}}O)-\Delta H_{c}^{o}$
$\Delta H_{c}^{0}(cyclopropane)=3\times (-393kJ/mol)+3\times (-285.8kJ/mol)-(-2091kJ/mol)$
The change in enthalpy when one mole of a substance at 1 atm and 298.15K is formed from its pure elements under the same conditions is known as the standard enthalpy of formation.

Hence, the enthalpy of formation of cyclopropane = -56.1kJ/mol

Note: For calculation of the heat of reaction from the heats of formation of the reactants and products from Hess’s law. Cyclopropene and propene have the same molecular formula, thereby producing the same combustion end products. The standard enthalpy of formation of pure elements in its reference form standard formation of enthalpy is zero. So, the standard enthalpy of formation of pure graphite, oxygen molecules, and hydrogen molecules will be zero.