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At 783 K in the reaction H(2)(g)+I(2)(g)...

At 783 K in the reaction `H_(2)(g)+I_(2)(g) hArr 2HI(g)`, the molar concentrations `("mol"^(-1))"of "H_(2),I_(2)and HI` at some instant of time are `0.1, 0.2 and 0.4`, respectively. If the equilibrium constant is 46 at the same temperature, then as the reaction proceeds

A

the amount of HI will increase

B

the amount of HI will decrease

C

the amount of `H_(2) and I_(2)` will increase

D

the amount of `H_(2) and I_(2)` will not change

Text Solution

Verified by Experts

The correct Answer is:
A
Reaction quotient
`Q=([HI]^(2))/([H_(2)][I_(2)])=(0.4xx0.4)/(0.1xx0.2)`
`Q=8`
`Q ltK`
So reaction will proceeds in forward direction. ltbr? Hence amount of HI increases.
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