Question

When ethyne is passed through a red hot tube, then formation of benzene takes place:

& \Delta {{H}_{f({{C}_{2}}{{H}_{2}})(g)}}=230kJmo{{l}^{-1}} \\\ & \Delta {{H}_{f({{C}_{6}}{{H}_{6}})(g)}}=85kJmo{{l}^{-1}} \\\ \end{aligned}$$ Calculate the standard heat of trimerisation of ethyne to benzene. ${{C}_{2}}{{H}_{2}}(g)\to {{C}_{6}}{{H}_{6}}(g)$ A. $205kJ\,mo{{l}^{-1}}$ B. $605kJ\,mo{{l}^{-1}}$ C. $-605kJ\,mo{{l}^{-1}}$ D. $-205kJ\,mo{{l}^{-1}}$

Answer 1

Based on the data given in the question, the equation can be formed which includes heat of trimerization, heat of ethyne and benzene. The data needs to be substituted and then we can obtain the required answer. Enthalpy can be both positive or negative.

Complete step by step answer:
- In order to answer our question, we need to learn about the enthalpy. We know that energy change occurring during the reaction at constant temperature and constant volume is given by internal energy change i.e., heat absorbed at constant volume is equal to change in the internal energy. However most of the reactions in the laboratory are carried out in open beakers or test tubes, etc. In such cases, the reacting system is open to the atmosphere. Since atmospheric pressure is almost constant, therefore, such reactions may involve change in volume. The energy change occurring during such reactions is not necessarily the same as the internal energy change.
- However, there are some rules when enthalpy of thermochemical equations are involved. For example, if a particular chemical equation has a standard enthalpy, then on multiplying or dividing the number of moles of the products and reactants by a constant, then the enthalpy also gets multiplied or divided by the same constant. When two or more reactions are added or subtracted, then the enthalpies also get added or subtracted accordingly. Now let us come to the solution. Let us write the reaction for trimerization:
$3{{C}_{2}}{{H}_{2}}\xrightarrow{trimerisation,\Delta }{{C}_{6}}{{H}_{6}}$
- We can see that 3 moles of ethyne are needed. So, by following the rules mentioned above, the standard heat of trimerization is:
$\Delta {{H}_{trimerisation}}=\Delta {{H}_{{{C}_{6}}{{H}_{6}}}}-3\times \Delta {{H}_{{{C}_{2}}{{H}_{2}}}}$
- We have multiplied by 3 as there are 3 moles present. So, the final answer is:
$\Delta {{H}_{trimerisation}} = 85 - 3(230)kJ\,mo{{l}^{-1}}$,
which gives $\Delta {{H}_{trimerisation}} = -605kJ\,mo{{l}^{-1}}$
The correct option is “C” .

Note: It is to be noted that when we talk about enthalpy then only the difference of enthalpy is taken $\Delta H$. It is so because absolute 0 temperature cannot be found practically, so the 0 point or origin for enthalpy cannot be found.