Naphthalene- a white solid used to make mothballs- has a vapour pressure of 0-10 mm Hg at 27- C- K p in atm and K c for the equilibrium are -C 10 H 8 s - C 10 H 8 g A- 3-4 - 10-2- 5 - 10-5- 5-34 - 10-6- 1-82 - 10-4- 1-32 - 10-4- 5-34 - 10-6D- 5 - 10-5- 3-4 - 10-2

The correct option is C 1.32 × 10^ 4, 5.34 × 10^ 6The given reaction is: nbsp; C_10H_8 (s) ⇌ C_10H_8 (g) Now, K_P = P_C_10H_8 = 0.10 mm of Hg As nbsp; we ...

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Standard XII

Chemistry

Derivation of Kp and Kc

Naphthalene, ...

Question

Naphthalene, a white solid used to make mothballs, has a vapour pressure of $0.10 m m H g$ at $27^{o} C$ .
$K_{p}$ (in atm) and $K_{c}$ for the equilibrium are :
$C_{10} H_{8} (s) ⇌ C_{10} H_{8} (g)$

5 \times 10^{- 5}, 3.4 \times 10^{- 2}

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3.4 \times 10^{- 2}, 5 \times 10^{- 5}

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

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5.34 \times 10^{- 6}, 1.82 \times 10^{- 4}

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Solution

The correct option is C $1.32 \times 10^{- 4}, 5.34 \times 10^{- 6}$
The given reaction is:-
$C_{10} H_{8} (s) ⇌ C_{10} H_{8} (g)$
Now, $K_{P} = P_{C_{10} H_{8}} = 0.10 m m o f H g$
As we know, 760 mm of Hg = 1 atm
$∴ 0.10 m m o f H g = \frac{0.1}{760} a t m = 1.32 \times 10^{- 4}$
$∴ K_{p} = 1.32 \times 10^{- 4} a t m$
$△ n_{g} = 1$
and $K_{P} = K_{C} (R T)^{△ n_{g}}$
$\Rightarrow K_{C} = \frac{K_{p}}{R T} = \frac{0.1}{760 \times R \times 300} = \frac{0.1}{760 \times 0.0821 \times 300} = 5.34 \times 10^{- 6}$

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Naphthalene, a white solid used to make mothballs, has a vapour pressure of 0.10 mm Hg at 27oC. Kp (in atm) and Kc for the equilibrium are : C10H8(s)⇌C10H8(g)

Naphthalene, a white solid used to make mothballs, has a vapour pressure of $0.10 m m H g$ at $27^{o} C$ .
$K_{p}$ (in atm) and $K_{c}$ for the equilibrium are :
$C_{10} H_{8} (s) ⇌ C_{10} H_{8} (g)$