Question

The combustion of methane can be represented as follows:

a. Use this information to determine the value of $\mathbf{∆}\mathit{H}$for the combustion of methane to form $\mathit{C}{\mathit{O}}_{\mathbf{2}}\left(g\right)$and $\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)\mathbf{.}$

b. What is$\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{o}}$ for an element in its standard state? Why is this? Use the figure above to support your answer.

c. How does for the reaction$\mathit{C}{\mathit{O}}_{\mathbf{2}}\left(g\right)\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\left(l\right)\stackrel{}{\mathbf{\to }}\mathit{C}{\mathit{H}}_{\mathbf{4}}\left(g\right)\mathbf{+}\mathbf{2}{\mathit{O}}_{\mathbf{2}}\left(g\right)$compare to that of the combustion of methane? Why is this?

Short answer

a. The value of for the combustion of methane is $-891 kJ$.

b. The $∆H_f^o$for any element in its standard state is zero.

c. The value of for the reverse reaction $C{O}_2\left(g\right)+2H_{2}O\left(l\right)\stackrel{}{\to }C{H}_4\left(g\right)+2O_2\left(g\right)$is $+891 kJ$.

Step-by-step solution

Calculate the enthalpy change for the combustion of methane.

The balanced chemical equation for the combustion of methane is shown below:

$C{H}_4\left(g\right)+2O_2\left(g\right)\stackrel{}{\to }C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

In step 1, given reaction (a) shows methane is transformed into its constitute elements (carbon and hydrogen).

$C{H}_4\left(g\right)\stackrel{}{\to }C\left(s\right)+2H_2\left(g\right)∆H_a=75 kJ$

In reaction (b), molecular oxygen is present in elemental form at standard state.

$O_2\left(g\right)\stackrel{}{\to }{O}_2\left(g\right)∆H_b=0 kJ$

Therefore, there is no changes in the reaction

In step 2, given reaction (c) represents the product formation in which carbon dioxide is formed from the elemental carbon and oxygen.

$C\left(s\right)+O_2\left(g\right)\stackrel{}{\to }C{O}_2\left(g\right)∆H_c=-394 kJ$

Similarly, reaction (d) is the formation reaction of water form its constitute elements such as hydrogen and oxygen as shown below:

$2H_2\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2H_{2}O\left(l\right)∆H_d=-572 kJ$

The enthalpy change for this reaction can be calculated by using the Hess’s law as shown below:

$$\begin{gathered} C{H}_4\left(g\right)\stackrel{}{\to }C\left(s\right)+2H_2\left(g\right)∆H_a=75 kJ \\ {O}_2\left(g\right)\stackrel{}{\to }{O}_2\left(g\right)∆H_b=0 kJ \\ C\left(s\right)+O_2\left(g\right)\stackrel{}{\to }C{O}_2\left(g\right)∆H_c=-394 kJ \\ 2H_2\left(g\right)+O_2\left(g\right)\stackrel{}{\to }2H_{2}O\left(l\right)∆H_d=-572 kJ \\ ----------------------- \\ C{H}_4\left(g\right)+2O_2\left(g\right)\stackrel{}{\to }C{O}_2\left(g\right)+2H_{2}O\left(l\right)∆H=? \end{gathered}$$

The enthalpy change for the overall reaction is the sum of the enthalpy change of the individual steps taken place.

$$\begin{gathered} ∆H=∆H_a+∆H_b+∆H_c+∆H_d \\ =75 kJ+0 kJ-394 kJ-572 kJ \\ =-891 kJ \end{gathered}$$

Hence, the value of$∆H$ for the combustion of methane is$-891 kJ$ .

Find the standard enthalpy of formation for any element in its standard state.

The $∆H_f^o$for any element in its standard state is zero because the absolute enthalpy value can not be determined. We know that the enthalpy is a state function. So, it does not depend on the path taken. Therefore, the elements in their standard state is assume be to zero as the reference point.

The chemical reaction is divided into two steps in order to determine the enthalpy change. The first step is the enthalpy change associated with the conversion of the elements (standard state) from the reactants. The second step is the enthalpy change associated with the conversion of the products from the elements.

The assumed reference point in both steps cancels out when the first and the second steps are added together. Finally, we obtain only the enthalpy change for the reaction. Therefore, the $∆H_f^o$for any element in its standard state is zero.

Calculate the enthalpy change for the reverse reaction.

The balanced chemical equation for the reaction is given as follow:

$C{O}_2\left(g\right)+2H_{2}O\left(l\right)\stackrel{}{\to }C{H}_4\left(g\right)+2O_2\left(g\right)$

In the above reverse reaction, the products in the combustion of methane is transformed into the elements in their standard state. The enthalpy change required for this process is $966 kJ$. The elements produced in the first step is converted into the reactants back. The enthalpy change for this conversion is $-75 kJ$.

The enthalpy change for the reverse reaction is the sum of two process,$966 kJ$ and $-75 kJ$as shown below.

$$\begin{gathered} ∆H=966 kJ-75 kJ \\ =+891 kJ \end{gathered}$$

Hence, the value of $∆H$for the reverse reaction is $+891 kJ$.