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Standard XII

Chemistry

Derivation of Kp and Kc

The unit of K...

Question

The unit of $K_{p}$ for the following reaction at equilibrium will be :
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

a t m^{- 1}

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a t m^{- 2}

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a t m^{+ 1}

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a t m^{+ 2}

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Solution

The correct option is B $a t m^{- 2}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

$K_{p} u n i t s : a t m^{Δ n_{g}}$
where,
$Δ n_{g} =$ No. of moles of gaseous products $-$ No. of moles of gaseous reactants
$Δ n_{g} = 2 - 4 = - 2$

So,here
$K_{p} u n i t s : a t m^{- 2}$

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Similar questions

Q. For the reaction,

2 N H_{3} (g) ⇌ N_{2} (g) + 3 H_{2} (g),

the units of

K_{p}

will be.

K_{p}

for a reaction at

25^{o} C

in 10 atm is 8.568. Then

K_{c}

for the reaction at

25^{o} C

will be:

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

Q. In a reaction mixture containing

H_{2}, N_{2}

and

{N H}_{3}

at partial pressure of

1

atm,

2

atm and

3

atm respectively, the value of

K_{p}

725 K

4.28 \times 10^{- 5} {a t m}^{- 2}

. In which direction the net reaction will go?

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

Q. A reaction mixture containing

H_{2}, N_{2}

and

N H_{3}

has partial pressures 2 atm, 1 atm and 3 atm respectively at 725 K. If the value of

K_{p}

for the reaction,

N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)}

4.28 \times 10^{- 5} a t m^{- 2}

at 725 K, in which direction the net reaction will go?

Q. At a temperature of

1000 K

, equilibrium constant

K_{c}

for the following reaction is

100 m o l^{2} L^{- 2}

N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g)

What will be the value of

K_{p}

for the reaction below?

N H_{3} (g) ⇋ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)

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Derivation of Kp and Kc

Standard XII Chemistry

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The unit of Kp for the following reaction at equilibrium will be : N2 (g)+3H2 (g)⇌2NH3 (g)

The correct option is B atm−2N2 (g)+3H2 (g)⇌2NH3 (g) Kp units: atmΔng where, Δng= No. of moles of gaseous products − No. of moles of gaseous reactants Δng=2−4=−2 So,here Kp units: atm−2

The unit of $K_{p}$ for the following reaction at equilibrium will be :
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

The correct option is B $a t m^{- 2}$
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$

$K_{p} u n i t s : a t m^{Δ n_{g}}$
where,
$Δ n_{g} =$ No. of moles of gaseous products $-$ No. of moles of gaseous reactants
$Δ n_{g} = 2 - 4 = - 2$

So,here
$K_{p} u n i t s : a t m^{- 2}$