The combustion of pentane, c5h12, occurs via the reaction
c5h12(g)+8o2(g)→5co2(g)+6h2o(g)
with heat of formation values given by the following table:
substance δh∘f
(kj/mol)
c5h12 (g) -119.9
co2(g) −393.5
h2o(g) −241.8
calculate the enthalpy for the combustion of 1 mole of pentane.

Answer : The enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as

The equation used to calculate enthalpy change is of a reaction is:  

The equilibrium reaction follows:

The equation for the enthalpy change of the above reaction is:

We are given:

Putting values in above equation, we get:

Therefore, the enthalpy for the combustion of 1 mole of pentane is -3024.8 kJ

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.

the answer is option a.