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For the process 3/2 A rarr B at 298 K, t...

For the process `3/2 A rarr B` at 298 K, `triangle G^(@)` is 163 kJ `mol^(-1)` The compostion of the reaction mixture is [B] =1 and [A] =10000 Predict the direction of the reaction and the relation between reaction quotient (Q) and the equilbrium constant (K)

A

forward direction because `Q gt K`

B

Reverse direction because `Q gt K`

C

Forward direction because `Q lt K`

D

It is a t equillibrium as Q=K

Text Solution

Verified by Experts

The correct Answer is:
C
`Q=("conc of product")^(n)/("conc. Of reactent" )^(n)=(B)/(A)`
Also `triangle G=2.303 RT log k`
`therefore logk=(163xx1000)/(2.303xx8.3141xx298)/(2.303xx8.3141xx298)-(163000)/(5705.8)`
=28.57
k `gt` Q means
Forward reaction is favoured with Q `lt` k
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