Question

Naphthalene, a white solid used to make mothballs, has a vapour pressure of $0.10mm$ $Hg$ at ${27}^{\circ }C$ Hence, $K_p$ and $K_C$ for the equilibrium are: $C_{10}{H}_8\left(s\right)\rightleftharpoons C_{10}{H}_8\left(g\right)$

• 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. $5.34\times {10}^{-6},1.32\times {10}^{-4}$

Answer 1

Answer: (C)

$1.32\times {10}^{-4},5.34\times {10}^{-6}$

Solution:

$C_{10}{H}_8\left(s\right)\rightleftharpoons C_{10}{H}_8\left(g\right)$

$K_p=P_{C_{10}{H}_8}=\frac{0.10}{760}atm$

$K_p=1.315\times {10}^{-4}atm$

$K_p=K_c(RT{)}^{\Delta n}$

Here,

$T=27+273=300K$

$\Delta n=1-0=1$

$K_c=\frac{Kp}{(RT{)}^{\Delta n}}$

$K_c=\frac{1.315\times {10}^{-4}}{(0.0821\times 300)}$

$K_c=\frac{1.315\times {10}^{-4}}{8.21\times 3}=\frac{131.5\times {10}^{-6}}{24.63}$

$K_c=5.339\times {10}^{-6}$

Hence the correct option: C