Question

Enthalpy of formation $A B_{(g)}$ from $A_{2(g)}$ and $B_{2(g)}$ is represented by $\Delta_{f} H(A B)$ in $kJ / mol$. Thus, electronegativity difference $(\Delta X)$ between $A$ and $B$ is related with $\Delta_{f} H(A B)$ as

• A. $-\Delta_{f} H=129.13 \times(\Delta X)$
• B. $\Delta_{f} H=129.13 \times(\Delta X)$
• C. $-\Delta_{f} H=129.13 \times(\Delta X)^{2}$
• D. $\Delta_{f} H=129.13 \times(\Delta X)^{2}$

Answer 1

Answer: (C)

$A_{2(g)}+B_{2(g)} \rightarrow A B_{(g)}$

As per Pauling's electronegativity difference, $\Delta X$ is proportional to the $\Delta_{f} H$ as

$\therefore \,\,\,\, X_{A}-X_{B}=0.088 \sqrt{\Delta_{f} H},\,\,\,\, \therefore \Delta_{f} H=\frac{(\Delta X)^{2}}{0.088 \times 0.088}$

$-\Delta_{f} H=129.13 \times(\Delta X)^{2}$

Negative sign shows that bond formation is exothermic process.