Question

Naphthalene, a white solid used to make moth balls has a vapour pressure of 0.10 mm Hg at 27${}^{\circ }\text{C}$. Hence $K_p\text{and}{K}_c$ for the equilibrium are:
${\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\left(\text{s}\right)\rightleftharpoons {\text{C}}_{\text{10}}{\text{H}}_{\text{8}}\left(\text{g}\right)$

• A. $0.10,0.10$
• B. $0.10,4.1\times {10}^{-3}$
• C. $1.32\times {10}^{-4},5.34\times {10}^{-6}$
• D. $5.36\times {10}^{-6},1.32\times {10}^{-4}$

Answer 1

The correct option is C $1.32\times {10}^{-4},5.34\times {10}^{-6}$

The given reaction is :-

$C_{10}{H}_{8\left(S\right)}\rightleftharpoons C_{10}{H}_{8\left(g\right)}$

Now, $K_P=P_{C_{10}{H}_8}=0.10mmofHg$

As we know, $760mmofHg=1atm$

$\therefore 0.10mmofHg=\frac{0.1}{760}atm$

$\therefore K_p=0.1mmofHg=\frac{0.1}{760}atm=1.32\times {10}^{-4}$

$△n=1$

& $K_P=K_C{\left(RT\right)}^{△n}$

$\Rightarrow K_C=\frac{K_p}{RT}=\frac{0.1}{760\times R\times 300}=\frac{0.1}{760\times 0.0821\times 300}=5.34\times {10}^{-6}$