For the homogeneous gas reaction at $600$ K,

$4 N H_{3 (g)} + 5 O_{2 (g)} ⇌ 4 N O_{(g)} + 6 H_{2} O_{(g)}$ ,

the equilibrium constant $K_{c}$ has the unit:

(m o l d m^{- 3})^{- 1}

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(m o l d m^{- 3})

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(m o l d m^{- 3})^{10}

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(m o l d m^{- 3})^{- 9}

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Solution

The correct option is B $(m o l d m^{- 3})$
For the reaction, $4 N H_{3 (g)} + 5 O_{2 (g)} ⇌ 4 N O_{(g)} + 6 H_{2} O_{(g)}$ ,

the change in the number of moles of reactants and products
$Δ n_{g} = (4 + 6) - (4 + 5) = 1$

The unit of the equilibrium constant $K_{c} = (C o n c e n t r a t i o n)^{Δ n} = (m o l d m^{- 3})^{1}$

Hence, option B is correct.

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Similar questions

Q. For the homogeneous reaction

4 N H_{3} (g) + 5 O_{2} (g) ⇋ 4 N O (g) + 6 H_{2} O (g)

the equilibrium constant

K_{c}

has the unit of:

For the homogeneous gas phase equilibrium $4 N H_{3} (g) + 5 O_{2} (g) \Leftrightarrow 4 N O (g) + 6 H_{2} O (g)$ the units of equilibrium constant $^{'} K_{C}^{'}$ are

Q. For the decomposition of

A B

600

K

, the following data were obtained:

$[A B] m o l {d m}^{- 3}$	Initial rate of decomposition of $A B$ in $m o l {d m}^{- 3} s^{- 1}$
$0.20$	$2.75 \times 10^{- 8}$
$0.40$	$11 \times 10^{- 8}$
$0.60$	$24.75 \times 10^{- 8}$

The order of the decomposition of

A B

is:

Q. Choose the correct expression of

K_{c}

for reaction,

4 N H_{3} (g) + 5 O_{2} (g) ⇌ 4 N O (g) + 6 H_{2} O (g)

For the homogeneous gas reaction at 600 K, 4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g), the equilibrium constant Kc has the unit:

The correct option is B (mol dm−3) For the reaction, 4NH3(g)+5O2(g)⇌4NO(g)+6H2O(g), the change in the number of moles of reactants and products Δng=(4+6)−(4+5)=1The unit of the equilibrium constant Kc=(Concentration)Δn=(mol dm−3)1Hence, option B is correct.

For the homogeneous gas reaction at $600$ K,

$4 N H_{3 (g)} + 5 O_{2 (g)} ⇌ 4 N O_{(g)} + 6 H_{2} O_{(g)}$ ,

the equilibrium constant $K_{c}$ has the unit: