Question

The standard heat of combustion of graphite carbon is $- 393.5KJmo{l^{ - 1}}$ . The standard enthalpy of $C{O_2}$ is:

$$A.{\text{ }} + 393.5KJmo{l^{ - 1}} \\\ B.{\text{ }} - 393.5KJmo{l^{ - 1}} \\\ C.{\text{ }} + 196.75KJmo{l^{ - 1}} \\\ D.{\text{ }} - 196.75KJmo{l^{ - 1}} \\\$$

Answer 1

In order to solve the given problem and to find out the standard enthalpy of carbon dioxide, first we will see the chemical reactions for the given case that is the combustion of graphite carbon and further; we will see the reaction for the formation of carbon dioxide. We will see the difference between the numbers of carbon dioxide molecules formed in each case if they are different then we will divide or multiply according with the constant multiple to the heat of combustion of the graphite.

Complete step by step answer:
First let us see both the reactions.
Let us see the first reaction which is the combustion of graphite.
$C\left( {{\text{graphite}}} \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right)$
For the given case we have the heat of combustion is:
$\Delta H = - 393.5KJmo{l^{ - 1}}$
Now let us see the reaction for the formation of carbon dioxide.
$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right)$
We can see that chemically both the reactions are the same with no difference even in the constant part multiple of the molecule.
So we do not need to even multiply or divide the above enthalpy for finding the standard enthalpy of carbon dioxide. And the standard enthalpy for carbon dioxide is $\Delta H = - 393.5KJmo{l^{ - 1}}$ .
Hence, the standard enthalpy of $C{O_2}$ is: $- 393.5KJmo{l^{ - 1}}$
So, the correct answer is “Option B”.

Additional Information:
Graphite is a crystalline shape with the atoms arranged in a hexagonal arrangement of the element carbon. It occurs naturally in this form and is, under normal conditions, the most stable source of carbon. It transforms into a diamond under high pressures and temperatures. The normal combustion enthalpy is the difference in enthalpy as 1 mole of a material burns (combines vigorously with oxygen) under standard conditions; it is often referred to as "heat of combustion".

Note: In order to solve such types of problems students must remember the different common and standard reactions. As in the above case we had to just find out the difference between the reactions it would not have been easier without knowing the reaction.