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The vapour pressure of mercury is 0.002 mm Hg at 27 ° C H g ( l ) ⇌ H g ( g ) The value of equilibrium constant K C is

Options

  1. A1.068 × 1 0 - 7 M
  2. B0.002 M
  3. C8.12 × 1 0 - 5 M
  4. D3.9 × 1 0 - 5 M

Correct answer

A. 1.068 × 1 0 - 7 M

Step-by-step solution

H g ( l ) ⇌ H g ( g ) Δ n g = 1 K p = 0.002 760 = 2.63 × 1 0 - 6 atm K p = K c ( R T ) Δ n K c = K p R T = 2.63 × 1 0 - 6 0.0821 × 300 = 1.068 × 1 0 - 7 M

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