Question

How can the standard enthalpy of formation of $CO$ be calculated?

Answer 1

Hint:‌The standard enthalpy of formation of any compound can be calculated by using more than one chemical equation whose value of enthalpy changes are known to us, and by rearrangement of those equations we can get the desired product. This is because Hess’s law says that the path of a chemical reaction does not change the total energy of a compound.After that, the enthalpy exchanges which occur in those equations could be used to get the exact enthalpy of formation of the desired compound.


Complete step by step answer:
The enthalpy of formation of any compound is the energy which is required for the formation of that compound. It is a thermodynamic quantity which is usually difficult to measure directly in real life. So, we will consider the enthalpy of combustion of a compound and constituent elements present in it.
Hess's law tells us that the total energy change taking place in a system during a reaction is not dependent on the route of the reaction. In order to find the heat of formation of carbon monoxide, we will consider a stepwise process. At first we will write the equation involving combustion of carbon. The equation can be expressed as,
\[{{C}_{(s)}}+\dfrac{1}{2}{{O}_{2}}_{(g)}\to C{{O}_{(g)}}\]
Here we can see that the coefficient of the oxygen gas on the left hand side of the equation is half, because the right hand side only has one oxygen atom, and so in order to balance out the equation we write half as a coefficient.
Now, since it is being mentioned in the question, that we need to calculate the standard value of enthalpy of formation, we will take the standard values of enthalpy of carbon dioxide.
\[{{C}_{(s)}}+{{O}_{2(g)}}\to C{{O}_{2(g)}}\]
Here in this reaction, as we can see we are burning carbon with oxygen in order to get carbon dioxide. The heat exchange in this process is \[\Delta {{H}_{c2}}=-394kJ\]. Now, if we further write the expression of burning of carbon monoxide. We get,
\[C{{O}_{(g)}}+\dfrac{1}{2}{{O}_{2(g)}}\to C{{O}_{2(g)}}\]
As we can see this equation expresses the burning of carbon monoxide, and the enthalpy exchange has the value \[\Delta {{H}_{c1}}=-283kJ\]. Since we want to know the enthalpy of formation of the carbon monoxide, we need to combine these two equations in such a way that the carbon monoxide is made as a product. After rearranging these two equation and cancelling out the common molecules we will see that we get the equation
\[{{C}_{(s)}}+\dfrac{1}{2}{{O}_{2}}_{(g)}\to C{{O}_{(g)}}\]
So, now the enthalpy of formation would become,
\[\Delta {{H}_{f}}-\Delta {{H}_{c1}}=~\Delta {{H}_{c2}}\]
Now, we will substitute the values of enthalpies in order to get the value of enthalpy of formation of carbon monoxide.
\[\Delta {{H}_{f}}-283=~-394\]

So, after solving this equation we get, \[\Delta {{H}_{f}}=-111.0kJ/mol\], which is the required answer.

Note:The formation of the organic or inorganic compounds which are expressed by chemical equations are hypothetical in nature. For instance, we cannot cut the oxygen in half as we write the coefficient of the oxygen gas as half, in the equation of formation of carbon monoxide.