For the reaction, `H_(2) + I_(2)hArr 2HI, K = 47.6` . If the initial number of moles of each reactant and product is 1 mole then at equilibrium

Options

A. `[I_(2)] = [H_(2)], [I_(2)] gt [HI]`
B. `[I_(2)] = [H_(2)], [I_(2)] lt [HI]`
C. `[I_(2)] lt [H_(2)], [I_(2)] = [HI]`
D. `[I_(2)] gt [H_(2)], [I_(2)] = [HI]`

Answer

B

Explanation

`{:(,H_(2),+,I_(2),hArr,2HI,,,,,),("Initial moles",,,,,1,,1,,,1),("Reacted",,,,,x,,x,,,x),("Produced",,,,,,,,,,2x),("At equilibrium",,,,,1-x,,1-x,,,1+2x):}`
`K = ((1+2x)^(2)//V^(2))/((1-x)(1-x)//V^(2))`
`K = ((1+2x)^(2))/((1-x)^(2)) = 47.6`
`((1+2x)/(1-x))^(2) = (6.9)^(2)`
`1+2x =6.9(1-x)`
`1+2x =6.9-6.9x`
`8.9x = 5.9`
`x = (5.9)/(8.9) = 0.66`