Question

Which element is more likely to dissociate into its monatomic form at \(3000 \mathrm{K}, \mathrm{H}_{2}\) or \(\mathrm{N}_{2} ?\) Why?

Short answer

Answer: H2 is more likely to dissociate into its monatomic form at 3000 K because it has a lower bond dissociation energy compared to N2.

Step-by-step solution

Understand the dissociation process

Before comparing the dissociation of H2 and N2, we need to know the type of reactions involved. For each diatomic molecule, the dissociation process involves breaking the molecular bond to produce two monatomic atoms: $$ H_2 \xrightarrow[3000\:K]{\Delta H_{H_2}} 2H $$ $$ N_2 \xrightarrow[3000\:K]{\Delta H_{N_2}} 2N $$ Here, \(\Delta H_{H_2}\) and \(\Delta H_{N_2}\) are the bond dissociation energies (enthalpy change) required to break the respective molecular bonds.

Find the bond dissociation energies of H2 and N2

To compare their likelihood of dissociation, we should analyze the bond dissociation energies of both molecules. The molecule with lower bond dissociation energy will be more likely to dissociate as it requires less energy to break the bond. Let's find the bond dissociation energies of H2 and N2. Bond dissociation energy of H2: $$ \Delta H_{H_2} = 436\,\text{kJ/mol} $$ Bond dissociation energy of N2: $$ \Delta H_{N_2} = 942\,\text{kJ/mol} $$

Compare the bond dissociation energies

Now that we have the bond dissociation energies for H2 and N2, we can compare them: $$ \Delta H_{H_2} = 436\,\text{kJ/mol} \;<\; \Delta H_{N_2} = 942\,\text{kJ/mol} $$ Since the bond dissociation energy for H2 is lower than that of N2, it requires less energy to break the bond in H2, making it more likely to dissociate at 3000 K.

Conclusion

At 3000 K, H2 is more likely to dissociate into its monatomic form because it has a lower bond dissociation energy compared to N2.