Question

In a crystal, at ${827}^{\circ }C$, one out of ${10}^{10}$ lattice site is found to be vacant, while in the same solid, one out of $2\times {10}^9$ lattice site is found to be vacant at ${927}^{\circ }C$. What is the enthalpy of vacancy formation in kJ/mol unit?

• A. $76.8$
• B. $176.8$
• C. $33$
• D. $23$

Answer 1

The correct option is B $176.8$

$\frac{{}^{n}V}{N}=A{e}^{-\Delta H_V/RT}⟶\left(x\right)$

${\left(\frac{{}^{n}V}{N}\right)}_{{827}^{o}C}=\frac{1}{{10}^{10}}⟶\left(1\right)$

${\left(\frac{{}^{n}V}{N}\right)}_{{927}^{o}C}=\frac{1}{2\times {10}^9}⟶\left(2\right)$ $\therefore T_1=827+273=1100K;T_2=927+273=1200K$

$\Rightarrow \frac{\left(2\right)}{\left(1\right)}=\frac{1/2\times {10}^9}{1/{10}^{10}}=\frac{{10}^{10}}{2\times {10}^9}=0.5\times 10=5$

$x\Rightarrow 2.303\log 5=-\frac{\Delta H_V}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

$\therefore \Delta H_V=2.303\times \log 5\times 8.314\times \left(\frac{1100\times 1200}{100}\right)$

$=1.766\times {10}^{5}J$

$=176.6kJ/mol$ $unit$ .