Solveeit Logo
Login

Question

Question: At 500 K, equilibrium constant, <img src="https://cdn.pureessence.tech/canvas_624.png?top_left_x=600...

At 500 K, equilibrium constant, for the following reaction is 5.

What would be the equilibrium constant for the reaction:

A

0.040.04

B

0.40.4

C

25

D

2.52.5

Answer

0.040.04

Explanation

Solution

: HI(g);Kc=5\mathrm { HI } _ { ( \mathrm { g } ) } ; \mathrm { K } _ { \mathrm { c } } = 5 …(i)

Multiply eqn (i) by 2,

H2( g)+I2( g)\mathrm { H } _ { 2 ( \mathrm {~g} ) } + \mathrm { I } _ { 2 ( \mathrm {~g} ) } ….(ii)

Now, reverse the reaction

H2( g)+I2( g);Kc=1(5)2\mathrm { H } _ { 2 ( \mathrm {~g} ) } + \mathrm { I } _ { 2 ( \mathrm {~g} ) } ; \mathrm { K } _ { \mathrm { c } } = \frac { 1 } { ( 5 ) ^ { 2 } }

∴Kc=125=0.04\therefore \mathrm { K } _ { \mathrm { c } } = \frac { 1 } { 25 } = 0.04