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Question: The standard molar heat of formation of ethane, \(C{O_2}\) and water are respectively \(( - 21.1),\,...

The standard molar heat of formation of ethane, CO2C{O_2} and water are respectively (−21.1),  (−94.1) and (−68.3)kcal( - 21.1),\,\,( - 94.1)\,and\,( - 68.3)kcal. The standard molar heat of combustion of ethane will be:
A. −372 kcal - 372\,kcal
B. 162 kcal162\,kcal
C. −240 kcal - 240\,kcal
D. 183.5 kcal183.5\,kcal

Explanation

Solution

We can solve the question by the information given above in this question. It is asked for the standard molar heat of combustion of ethane after which carbon dioxide and water are released. Write the balanced chemical equation firstly and then write the molar enthalpy for reaction. Molar enthalpy of reaction is written as the difference of enthalpies of products minus for the reactants.

Complete step-by-step answer:
As we have ethane and its combustion is done, it means ethane is reacting with full supply of oxygen after which we get product as carbon dioxide and water. The products are obtained the same when any hydrocarbon is burned in full presence of oxygen.
We have to write the complete balanced equation for above reaction,
C2H6 (g)+ 72O2(g) → 2CO2(g) + 3H2O(l){C_2}{H_6}\,(g) + \,\dfrac{7}{2}{O_2}(g)\, \to \,2C{O_2}(g)\, + \,3{H_2}O(l)
Here, for this balanced chemical reaction we can write the molar enthalpy of reaction as,
ΔHC= (2ΔHf(CO2)+3ΔHf(H2O))− (ΔHf(C2H6) + 72ΔHf(O2))\Delta {H_C} = \,\left( {2\Delta {H_{f(C{O_2})}} + 3\Delta {H_{f({H_2}O)}}} \right) - \,\left( {\Delta {H_{f({C_2}{H_6})}}\, + \,\dfrac{7}{2}\Delta {H_{f({O_2})}}} \right)
On putting the values of enthalpies given above we can easily calculate the value of standard molar heat of combustion of ethane.
ΔHC= (2×−94.1+3×−68.3)− (−21.1 + 72×0)\Delta {H_C} = \,\left( {2 \times - 94.1 + 3 \times - 68.3} \right) - \,\left( { - 21.1\, + \,\dfrac{7}{2} \times 0} \right)
ΔHC= (−188.2−204.9)+ (42.2)= 350.9 k cal mol−1\Delta {H_C} = \,\left( { - 188.2 - 204.9} \right) + \,\left( {42.2} \right) = \,350.9\,k\,cal\,mo{l^{ - 1}}
As the standard molar enthalpy is 350.9 k cal mol−1350.9\,k\,cal\,mo{l^{ - 1}} the answer is closest to this is option A.

Therefore, Option (A) is correct.

Note: The standard molar enthalpy of any reaction is written as the sum of enthalpy of product subtracting the sum of enthalpy of the reactant. In the above expression we wrote the molar enthalpy of oxygen as zero, this is because the molar enthalpy of molecules which are in their reference state is always written as zero.