For the given hypothetical reactions, the equilibrium constants are as follows:

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
For the given hypothetical reactions, the equilibrium constants are as follows:

$X ⇌ Y; K_{1} = 1.0$

$Y ⇌ Z; K_{2} = 2.0$

$Z ⇌ W; K_{3} = 4.0$

The equilibrium constant for the reaction $X ⇌ W$ is [2024]

1 12.0

2 6.0

3 8.0

4 7.0

Ans.
(3)

On adding reactions, equilibrium constants are multiplied

   $X ⇌ Y, K_{1} = \frac{[Y]}{[X]}$

   $Y ⇌ Z, K_{2} = \frac{[Z]}{[Y]}$

   $Z ⇌ W, K_{3} = \frac{[W]}{[Z]}$

___________________

  Adding all three:

$X ⇌ W, K = \frac{[W]}{[X]} = \frac{[W] [Z] [Y]}{[Z] [Y] [X]} = K_{1} K_{2} K_{3} = 1 \times 2 \times 4 = 8$

Want Us to Email you About Special Offers & Updates?