`SO_(3) (g)` is produced as : `SO_(2) (g) + (1)/(2) O_(2) (g) hArr SO_(3) (g)` At `900 K, 0.2 `mol of `SO_(2)` and 0.4 mol of `O_(2)` are taken in 2L

Question

`SO_(3) (g)` is produced as : `SO_(2) (g) + (1)/(2) O_(2) (g) hArr SO_(3) (g)` At `900 K, 0.2 `mol of `SO_(2)` and 0.4 mol of `O_(2)` are taken in 2L

`SO_(3) (g)` is produced as :
`SO_(2) (g) + (1)/(2) O_(2) (g) hArr SO_(3) (g)`
At `900 K, 0.2 `mol of `SO_(2)` and 0.4 mol of `O_(2)` are taken in 2L vesset. When equilibrium is reaches by concentration of `SO_(3) (g)` is 0.08 M. Then `K_(C)^(0)` for reaction is :
A. `(10)/(sqrt2)`
B. `10^(-1)`
C. `10`
D. `100`

Monjur Alahi

10 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
`{:(,SO_(2) (g),+,(1)/(2)O_(2) (g),hArr,SO_(3) (g),T = 900 K,),("Concentration",(0.2)/(2),,(0.4)/(2),,,,),("initially",,,,,,,),("concentration",0.1 - x,,0.2 - (1)/(2) x,,x,,),(,x = 0.08 M,,,,,,),(,,,,,,,):}`
`K_(C) = ([SO_(3)])/([SO_(2)] [O_(2)]^(1//2)) = (0.08)/(sqrt(0.16) xx 0.02) = 10`

10 views Answered 2023-02-05 15:29