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Naphthalene, a white solid used to make mothballs, has a vapor pressure of 0.10 mm Hg at 27ºC. Hence, [tex]K_p[/tex] and [tex]K_c[/tex] for the equilibrium are:

\[ \text{C}_{10}\text{H}_8(s) \rightleftharpoons \text{C}_{10}\text{H}_8(g) \]

A. 0.10, 1.10
B. 0.10, 4.1 × 10⁻³
C. 1.32 × 10⁻⁴, 10⁻³
D. 5.34 × 10⁻⁶, 1.32 × 10⁻⁴

Answer :

The vapour pressure of naphthalene at 27ºC is 0.10 mm Hg. Therefore, the values of Kp and Kc for the equilibrium are 0.10, 0.10.

Explanation:

The vapour pressure of naphthalene at 27ºC is 0.10 mm Hg. The equilibrium expression for the vaporization of naphthalene can be written as:

Kp = P(gas) / P(solid) = 0.10

The equilibrium constant Kc can be related to Kp using the ideal gas law:

Kp = Kc * (RT)Δn

where Δn is the difference in moles between the gaseous and solid states. Since there is only one mole of naphthalene in both states, Δn = 0. Therefore, Kp = Kc * (RT)0 = Kc. Therefore, the values of Kp and Kc for the equilibrium are 0.10, 0.10.

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