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The equilibrium constant of the equilibr...

The equilibrium constant of the equilibrium equation
`H_2O(g)+CO(g)hArrH_2(g)+CO_2(g)`
is `0.44` at `1259K`. The value of equilibrium constant for the equilibrium equation
`H_2(g)+2CO_2(g)hArrH_2O(g)+CO(g)`
will be

A

`-0.44`

B

`-1//0.44`

C

`1//0.44`

D

`0.44`

Text Solution

Verified by Experts

The correct Answer is:
C
`K_(eq)^'` the equilibrium constant for the reaction written in reverse, is the reciprocal of `K_(eq)`, the equilibrium constant for the original reaction:
`K_(eq)^'=(1)/(K_(eq))=(1)/(0.44)`
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