Question

Estimate the enthalpy change for this reaction. Start by drawing the Lewis electron dot diagrams for each substance. \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}\)

Short answer

The estimated enthalpy change is \(-93\) kJ/mol.

Step-by-step solution

Draw Lewis Structures

Begin by drawing the Lewis structures for each of the substances involved in the reaction. For \(N_2\), two nitrogen atoms are triple bonded to each other, represented as :N≡N:. For \(H_2\), two hydrogen atoms share a single bond, represented as H-H. For \(NH_3\), each nitrogen atom forms three single bonds with three hydrogen atoms, represented as :H-N-H: and a pair of nonbonding electrons on nitrogen.

Identify and List the Bonds

List all the bonds present in the reactants and products. In \(N_2\), there is 1 triple bond between nitrogen atoms. In \(H_2\), there are 3 single hydrogen-hydrogen bonds for the three \(H_2\) molecules. In \(NH_3\), each of the 2 ammonia molecules contains 3 nitrogen-hydrogen single bonds, totaling 6 \(N-H\) bonds.

Calculate Energy of Bonds Broken

The enthalpy change is estimated by considering the energy required to break bonds minus the energy released when new bonds are formed. Start by calculating the total energy of bonds broken. For the \(1\) \(N\equiv N\) bond, use a bond energy of approximately 945 kJ/mol. For the \(3\) \(H-H\) bonds, use a bond energy of approximately 436 kJ/mol each. Total energy for breaking = \(1\times 945 + 3\times 436 = 945 + 1308 = 2253\) kJ.

Calculate Energy of Bonds Formed

Calculate the energy released by the formation of new bonds. There are 6 \(N-H\) bonds formed, each with an approximate bond energy of 391 kJ/mol. Total energy for forming = \(6\times 391 = 2346\) kJ.

Determine Enthalpy Change

Calculate the total enthalpy change (\(\Delta H\)) for the reaction using the formula \(\Delta H = \text{energy of bonds broken} - \text{energy of bonds formed}\). Substitute the values: \(\Delta H = 2253 - 2346 = -93\) kJ/mol. Therefore, the estimated enthalpy change for the reaction is \(-93\) kJ/mol.

Key concepts

Lewis structures

Lewis structures are diagrams that represent the covalent bonds between atoms in a molecule, along with any lone pairs of electrons. These structures are essential for understanding the arrangement of atoms and predicting molecular geometry.
For example, in the reaction \(_2 + 3 \h_2 ightarrow 2 h_3\) , we need to draw the Lewis structures of each molecule. \(_2\) is depicted as :N≡N: with a triple bond between the two nitrogen atoms, each with non-bonding pairs. \(h_3\) has a central nitrogen atom single-bonded to three hydrogen atoms, surrounded by a lone pair of electrons, forming the structure :H-N-H: . These diagrams provide a visual representation of how atoms are bonded and where potential energy is stored within the molecules.

bond energy

Bond energy refers to the amount of energy required to break one mole of a particular bond in a gaseous molecule. Bond energies are used to estimate the enthalpy change for chemical reactions by calculating the energy needed to break bonds in the reactants and the energy released from bonds formed in the products.
In this reaction, breaking the triple bond of \(_2\), which has a high bond energy of 945 kJ/mol, requires significant energy. Similarly, breaking the \(-h\) bonds, which are 436 kJ/mol each, in \(_2\), demands energy. Conversely, when forming six new single \(-h\) bonds in \(h_3\) at 391 kJ/mol each, energy is released, contributing to the enthalpy change calculation.

enthalpy calculation

Enthalpy calculation helps us determine whether a reaction absorbs or releases energy. It is calculated by subtracting the energy of bonds formed from the energy of bonds broken. This approach is based on Hess's law, implying that the enthalpy change of a reaction can be estimated using the sum of the enthalpy changes of individual steps.
To calculate for the reaction \(_2 + 3 \h_2 ightarrow 2 h_3\), first find the total energy required to break the bonds in the reactants, which amounts to 2253 kJ. Next, find the energy released by forming the \(-h\) bonds in \(h_3\), totaling 2346 kJ. The enthalpy change \(\Delta H\) is the difference: \(\Delta H = 2253 \text{kJ} - 2346 \text{kJ} = -93 \text{kJ/mol}\). Meaning, the reaction releases energy.

chemical reactions

Chemical reactions involve the rearrangement of atoms to transform reactants into products. These processes are governed by the breaking and formation of chemical bonds. In our specific reaction, \(_2 + 3 \h_2 ightarrow 2 h_3\), the nitrogen and hydrogen molecules react to form ammonia.
As the reaction progresses, the strong triple bond in nitrogen molecules and single bonds in hydrogen molecules break. These processes consume energy. New stable ammonia molecules form with nitrogen and hydrogen atoms bonded. The formation of these \(-h\) bonds releases energy. An understanding of these bond transformations is crucial to predict the energy changes associated with the reaction, as seen in the calculated enthalpy change.