Choose the correct unit of equilibrium constant $(K_{c})$ for the given reaction.
$C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$

mol/L

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{(mol/L)}^{\frac{3}{2}}

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(mol/L)^{\frac{- 1}{2}}

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(mol/L)^{\frac{1}{2}}

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Solution

The correct option is D $(mol/L)^{\frac{1}{2}}$
Equilibrium constant $K_{c}$ is proportional to the active masses of the products and reactants.
$C (s) + \frac{1}{2} O_{2} (g) \to C O (g)$
$For this reaction, K_{c} = \frac{[C O]}{[O_{2}]^{\frac{1}{2}}}$
Active mass of solids are considered as $1$ .
where $[C O]$ , $[O_{2}]^{\frac{1}{2}}$ are the molar concentations having a unit of mol/L
So, unit of $K_{c} = \frac{m o l / L}{(m o l / L)^{\frac{1}{2}}}$
$K_{c} = (m o l / L)^{\frac{1}{2}}$

Suggest Corrections

Similar questions

(K_{c})

C (s) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)

(K_{c})

C (s) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)

(K_{c})

C (s) + \frac{1}{2} O_{2} (g) ⇌ C O (g)

C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H_{1} = a

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H_{2} = b

C (s) + C O_{2} (g) \to 2 C O (g); Δ H_{3} = c

^{\circ} C

Z n_{(s)} + 1 / 2 O_{2 (g)} ⟶ Z n O_{(s)}, Δ G^{\circ} = - 360 k J m o l^{- 1}

C_{(s)} + 1 / 2 O_{2 (g)} ⟶ C O_{(g)}, Δ G^{\circ} = - 460 k J m o l^{- 1}

^{\circ} C

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