Bond Dissociation Enthalpy — MCQ Practice

Which of the following C-H bonds would have the lowest Bond Dissociation Enthalpy (BDE)?

Given the following average bond enthalpies: C-H = $413,\text{kJ/mol}$ Cl-Cl = $242,\text{kJ/mol}$ C-Cl = $328,\text{kJ/mol}$ H-Cl = $431,\text{kJ/mol}$ Calculate the enthalpy change ($Delta H$) for the reaction: $CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)$.

Which of the following statements about Bond Dissociation Enthalpy (BDE) is INCORRECT?

Arrange the following C-H bonds in increasing order of their Bond Dissociation Enthalpy (BDE): I. C-H in $CH_4$ II. C-H in $CH_3CH_2CH_3$ (secondary C-H) III. C-H in $CH_2=CH-CH_2-H$ (allylic C-H) IV. C-H in $CH_3-CH(CH_3)-CH_3$ (tertiary C-H)

If the enthalpy of formation of $H_2O(g)$ is $-242,\text{kJ/mol}$, and the bond dissociation enthalpies are $D(H-H) = 436,\text{kJ/mol}$ and $D(O=O) = 498,\text{kJ/mol}$, what is the average bond enthalpy of the O-H bond in $H_2O(g)$?

Which of the following factors would generally lead to a *higher* Bond Dissociation Enthalpy (BDE) for a given bond?

The BDE of the C-H bond in $CH_4$ is approximately $439,\text{kJ/mol}$. The BDE of the C-H bond in $CH_3^cdot$ (i.e., to form $CH_2^cdot + H^cdot$) is approximately $460,\text{kJ/mol}$. Why is there a difference in these values?