Solveeit Logo

Question

Question: When ethyne is passed through a red hot tube, then benzene formation takes place? Given, \( {H^o}_...

When ethyne is passed through a red hot tube, then benzene formation takes place?
Given, Hof(C2H2)(g)=230kJ mol1{H^o}_{f({C_2}{H_{2)(g)}}} = 230kJ{\text{ mo}}{{\text{l}}^{ - 1}} ,
ΔHof(C6H6)(g)=85kJ mol1\Delta {H^o}_{f({C_6}{H_{6)(g)}}} = 85kJ{\text{ mo}}{{\text{l}}^{ - 1}}
Calculate the standard heat of trimerisation of ethyne to benzene?
3C2H2(g)C6H6(g)3{C_2}{H_2}(g) \to {C_6}{H_6}(g)
(A) 205 kJ mol1{\text{205 kJ mo}}{{\text{l}}^{ - 1}}
(B) 605 kJ mol1{\text{605 kJ mo}}{{\text{l}}^{ - 1}}
(C)  - 605 kJ mol1{\text{ - 605 kJ mo}}{{\text{l}}^{ - 1}}
(D)  - 205 kJ mol1{\text{ - 205 kJ mo}}{{\text{l}}^{ - 1}}

Explanation

Solution

Hint : As in the question, standard heat of formation of benzene from ethyne is asked, so to solve this question, first we need to understand the enthalpy of formation or the enthalpy change during the formation of products from reactants.

Complete Step By Step Answer:
So, the standard enthalpy of formation is the enthalpy change when 1mol1mol of a product is formed at standard state (or in standard conditions) from the pure reactant at the same conditions.
Now, the standard conditions are - 1atm1atm pressure and 298.15K298.15K temperature.
Also we know the formula to calculate the standard heat of formation-
ΔHof(reaction)=ΔHof(products)ΔHof(reactants)\Delta {H^o}_{f(reaction)} = \sum {\Delta {H^o}_f(products) - \sum {\Delta {H^o}_f({\text{reactants)}}} }
In our case, equation given is
3C2H2(g)C6H6(g)3{C_2}{H_2}(g) \to {C_6}{H_6}(g)
So, benzene is our product and ethyne is the reactant.
Hence, putting values in the formula to calculate standard heat of formation, we get –
ΔHof(reaction)=ΔHof(products)ΔHof(reactants)\Delta {H^o}_{f(reaction)} = \sum {\Delta {H^o}_f(products) - \sum {\Delta {H^o}_f({\text{reactants)}}} }
ΔHof(3C2H2(g)C6H6(g))=ΔHof(C6H6)(g))ΔHof((C2H2)(g))\Delta {H^o}_{f(3{C_2}{H_2}(g) \to {C_6}{H_6}(g))} = \sum {\Delta {H^o}_f{(_{{C_6}{H_{6)(g)}}}}) - \sum {\Delta {H^o}_f{(_{({C_2}{H_{2)(g)}}}}{\text{)}}} }
ΔHotrimerisation=85(3×230)\Delta {H^o}_{trimerisation} = 85 - (3 \times 230)
 = - 605 kJ mol1{\text{ = - 605 kJ mo}}{{\text{l}}^{ - 1}}
Thus, by solving the equation, we get our answer  = - 605 kJ mol1{\text{ = - 605 kJ mo}}{{\text{l}}^{ - 1}}
Thus option C is the correct answer.

Note :
In other words, we can also say that the enthalpy change is the amount of heat absorbed or released during a reaction of conversion of reactants into its products at the standard pressure and temperature. The change in enthalpy is denoted by the symbol Hof{H^o}_f .
There is one other formula through which our change in enthalpy can be calculated and that is ΔH=msΔT\Delta H = ms\Delta T , where m is the mass of the substance and s is the specific heat and ΔT\Delta T is change in temperature.