Question

Question: (a) Draw a Lewis diagram for carbonic acid,${\mathbf{\text{H}}}_{\mathbf{2}}{\mathbf{\text{CO}}}_{\mathbf{3}}$ , with a central carbon atom bonded to the three oxygen atoms.

(b) Carbonic acid is unstable in aqueous solution and converts to dissolved carbon dioxide. Use bond enthalpies to estimate the enthalpy change for the following reaction:

${\mathbf{\text{H}}}_{\mathbf{2}}{\mathbf{\text{CO}}}_{\mathbf{3}}\mathbf{\to }{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{+}{\mathbf{\text{CO}}}_{\mathbf{2}}$

Short answer

( a ) The Lewis diagram for carbonic acid is shown below.

( b ) The change in enthalpy is $26kJ/\text{mol}$.

Step-by-step solution

Given information

The given reaction is ${\text{H}}_2{\text{CO}}_3\to {\text{H}}_2\text{O}+{\text{CO}}_2$.

Concept of Lewis diagram

Lewis structures are diagrams that show the bonding between the atoms in a molecule as well as the lone electron pairs. The total number of electrons must be equal to the sum of the numbers of valence electrons on each individual atom. The lines between the atoms represent the shared electron pair.

Draw the Lewis diagram of carbonic acid

(a)

The number of valence electrons for carbon oxygen and hydrogen is 4, 6 ,and 1.

Calculation of enthalpy change

(b)

The reaction is written in Lewis structure:

The enthalpy for $\text{C}-\text{O}$,$436kJ/mole$,$\text{C}=\text{O}$and $H-\text{H}$bond is $351kJ/mole$, $728kJ/mole$, $463kJ/mole$and $436kJ/mole$.

The enthalpy for the reactants and products is calculated as:

$\begin{array}{l}{\text{H}}_2{\text{CO}}_3\\ \text{O}-\text{Hbond:}2\to 926\text{kJ}/\text{mol}\\ \text{C}-\text{Obond}:2\to 702\text{kJ}/\text{mol}\\ \text{C}=\text{Obond}:1\to 728\text{kJ}/\text{mol}\end{array}$

So, total enthalpy$=2356\text{kJ}/\text{mol}$.

For ${\text{CO}}_2$:

$\text{C}=\text{O}$ bonds $:2\to 1456\text{kJ}/\text{mol}$

Total enthalpy $=1456\text{kJ}/\text{mol}$.

For ${\mathbf{H}}_2\mathbf{O}$:

$\text{O}-\text{H}$ bonds $:2\to 926\text{kJ}/\text{mol}$

So, total enthalpy$=926\text{kJ}/\text{mol}$.

So, the change in enthalpy is given as:

data-custom-editor="chemistry" $\begin{array}{l}\Delta H=\Delta H_{C{O}_2}+\Delta H_{H_{2}O}-\Delta H_{H_{2}C{O}_3}\\ \Delta H=1456+926-2356\\ \Delta H=26k\text{k}/\text{mol}\end{array}$