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Question: When ethyne is passed through a red hot tube, then formation of benzene takes place. Calculate the s...

When ethyne is passed through a red hot tube, then formation of benzene takes place. Calculate the standard heat of trimerisation of ethyne to benzene.
C2H2(g)→C6H6(g){{C}_{2}}{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(g)
Given: ΔH∘f(C2H2)(g)=230kJ⁡mol⁡−1\Delta {{H}^{{}^\circ }}_{f({{C}_{2}}{{H}_{2}})(g)}=230\operatorname{kJ}{{\operatorname{mol}}^{-1}}and ΔH∘f(C6H6)(g)=85kJ⁡mol⁡−1\Delta {{H}^{{}^\circ }}_{f({{C}_{6}}{{H}_{6}})(g)}=85\operatorname{kJ}{{\operatorname{mol}}^{-1}}

Explanation

Solution

ΔH∘f\Delta {{H}^{{}^\circ }}_{f} is known as the standard enthalpy change of formation. Use the equation given below to find the answer:
ΔH∘(reaction)=∑ΔH∘f(products)−∑ΔH∘f(reactants)\Delta {{H}^{{}^\circ }}_{(reaction)}=\sum \Delta {{H}^{{}^\circ }}_{f(products)}-\sum \Delta {{H}^{{}^\circ }}_{f(reactants)}

Complete answer:
Let us first discuss enthalpy. It is defined as the amount of heat energy in a system. Enthalpy is calculated as the sum of the internal energy (U) with the product of pressure and volume of a system.

Mathematically it is represented as:
H=U+PVH=U+PV
ΔH\Delta H is the change in enthalpy over a specified period of time. ΔH∘\Delta {{H}^{{}^\circ }} is the standard change in enthalpy over a period of time. So now, we can come toΔH∘f\Delta {{H}^{{}^\circ }}_{f}; the subscript “f” denotes “formation”. SoΔH∘f\Delta {{H}^{{}^\circ }}_{f} is the standard enthalpy change of formation of a specific compound in a given chemical reaction.
Remember that, enthalpy is a state function, which means, the change in enthalpy in a given process will only depend on its initial and final stages and not how the process occurred. That means, in a given chemical reaction, the enthalpy change can be simply calculated if we take the enthalpy of formation of all the reactants and products into account. That is what we are going to ask in this question.
The above theory has a specific formula which has already been mentioned in the hint. It is as follows:
ΔH∘(reaction)=∑ΔH∘f(products)−∑ΔH∘f(reactants)\Delta {{H}^{{}^\circ }}_{(reaction)}=\sum \Delta {{H}^{{}^\circ }}_{f(products)}-\sum \Delta {{H}^{{}^\circ }}_{f(reactants)}

In this case, the reactant is ethyne molecule and the product is benzene molecule. According to the stoichiometric coefficients in the chemical reaction, three molecules of ethyne react to form one molecule of benzene. Therefore, the standard change in enthalpy of formation of ethyne will be multiplied by three while that of benzene will be multiplied by one. The equation is as follows:

ΔH∘(reaction)=1×ΔH∘f(benzene)−3×ΔH∘f(ethyne)\Delta {{H}^{{}^\circ }}_{(reaction)}=1\times \Delta {{H}^{{}^\circ }}_{f(benzene)}-3\times \Delta {{H}^{{}^\circ }}_{f(ethyne)}

Putting the respective values in the above equation we get:

& \Delta {{H}^{{}^\circ }}_{(reaction)}=85-3\times 230 \\\ & \Delta {{H}^{{}^\circ }}_{(reaction)}=-605\operatorname{kJ}{{\operatorname{mol}}^{-1}} \\\ \end{aligned}$$ Therefore, the standard heat of trimerisation, which is the same as the standard enthalpy change of reaction is $605\operatorname{kJ}{{\operatorname{mol}}^{-1}}$. **Note:** Do not forget to take the stoichiometric coefficients into account while calculating the standard enthalpy change of reaction. The$\Delta {{H}^{{}^\circ }}$in this case is negative, which means the reaction is exothermic. That is, the system taken into consideration releases heat into the surrounding.