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Question

On reversing a reaction equilibrium $(Δ n = 0)$ , the numerical value of $K_{c}$ and its unit both are reversed.
True
False

A

True

B

False

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Solution

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For a reaction $a A + b B ⇌ c C + d D$ ,

$K = \frac{C^{c} D^{d}}{A^{a} B^{b}}$

For, $c C + d D ⇌ a A + b B$ ,

$K = \frac{A^{a} B^{b}}{C^{c} D^{d}}$

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Similar Questions

Q1

On reversing a reaction equilibrium

(Δ n = 0)

, the numerical value of

K_{c}

and its unit both are reversed.

View Solution

Q2

On reversing a reaction equilibrium

(Δ n \neq 0)

, the numerical value of

K_{c}

and its units both are reversed.
If true enter 1, if false enter 0.

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Q3

(A) In a gaseous reaction,

K_{c}

is unitless when

Δ n = 0

.
(R) Unit of

K_{c} = (m o l L^{- 1})

.

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Q4

S :

K_{c}

,

K_{p}

and

K_{x}

are the equilibrium constants of a reaction in terms of concentration, pressure and mole fraction respectively.
E :

K_{c}

and

K_{p}

do not depend on equilibrium pressure but

K_{x}

depends upon equilibrium pressure if

Δ n \neq 0

.

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Q5

For a reversible reaction

\frac{K_{p}}{K_{c}} = {(R T)}^{Δ n}

at

300 K

. If

\frac{K_{p}}{K_{c}} = 14941.5

(in litre atm unit) for a given reaction then

Δ n

for the reaction is :

[Given that

l o g \frac{K_{p}}{K_{c}} = 4.17

and

l o g (24.93) = 1.39

]

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