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If the equilibrium constant of the rever...

If the equilibrium constant of the reversible reaction `HI(g)hArr1//2H_2(g)+1//2I_2(g)` is `7.4`, the equilibrium constant for the reversible reaction `2HI(g)hArrH_2(g)+I_2(g)` will be

A

`sqrt(7.4)`

B

`54.76`

C

`14.8`

D

`7.4`

Text Solution

Verified by Experts

The correct Answer is:
B
Note that the second equation is obtained when we multiply the first equation by 2. Always remember that if an equation for a reaction is multiplied by any factor n, then the original value of `K_(eq)` is raised to the nth power, i.e.,
`K_(eq)^'=(K_eq)^n`
Here, `n=2`. Thus,
`K_(eq)^'=(7.4)^2=54.76`
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