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Introduction

Consider the ...

Question

Consider the following reaction
$A + B ⇌ E; K_{c} = 6$
$2 B + C ⇌ 2 D; K_{c} = 4$
What will be the equilibrium constant ( $K_{c}$ ) for the following reaction?
$A + D ⇌ E + \frac{1}{2} C$

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Solution

Consider the following reaction
$A + B ⇌ E; K_{c} = 6$ ......(1)
$2 B + C ⇌ 2 D; K_{c} = 4$ .....(2)
The equilibrium constant ( $K_{c}$ ) for the following reaction will be $3$
$A + D ⇌ E + \frac{1}{2} C$
Reverse reaction (2)
$2 D ⇌ 2 B + C; K_{c} = \frac{1}{4} = 0.25$ .....(4)
Divide the reaction (4) with 2
$D ⇌ B + \frac{1}{2} C; K_{c} = \sqrt{0.25} = 0.5$ .....(5)
Add reactions (1) and (5)
$A + D ⇌ E + \frac{1}{2} C; K_{c} = 6 \times 0.5 = 3$

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Similar questions

If $A + B ⇌ C + D$ ; $K_{C} = K_{1}$ and $E^{⊖} = a V$ ; $2 A + 2 B ⇌ 2 C + 2 D$ ; $K_{C} = K_{2}$ and $E^{⊖} = b V$ then:

2 A + 2 B ⇌ 2 C + 2 D

K_{c} = 1 / 6

K_{c}

C + D ⇌ A + B

K_{c}

A + B ⇋ C

K_{c}

2 A + D ⇋ C

K_{c}

C + D ⇋ 2 B

K_{c}

A + B \Leftrightarrow C

4

K_{c}

2 A + D \Leftrightarrow C

6

K_{c}

C + D \Leftrightarrow 2 B

A ⇌ B; K_{c} = 2,

B ⇌ C; K_{c} = 4,

C ⇌ D; K_{c} = 6

K_{c}

A ⇌ D

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