For a particular isomer of C 8 H 18 , the combustion reaction produces 5099.5 kJ of heat per mole of C 8 H 18 ( g ) consumed, under standard conditions. C 8 H 18 ( g ) + 25 2 O 2 ( g ) ⟶ 8 CO 2 ( g ) + 9 H 2 O ( g ) Δ H ∘ rxn = − 5099.5 kJ / mol What is the standard enthalpy of formation of this isomer of C 8 H 18 ( g ) ?

the standard enthalpy of formation of this isomer of C₈H₁₈ (g) =  -375 kj/mol

Explanation:

The given combustion reaction

C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O   (1)

Heat of reaction or enthalpy of combustion = -5099.5 kj/mol

from equation (1)

     ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation                    of reactants)

Or,     - 5099.5   = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)](2)

Given ΔH⁰f(CO₂) = - 393.5 kj/mol  & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0

Using equation (2)

ΔH⁰f(C₈H₁₈) = -621 kj/mol

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.