If a mixture 0.4 mole `H_2` and 0.2 mole `Br_2` is heated at 700 K at equilibrium, the value of equilibrium constant is `0.25xx10^10` then find out th

Question

If a mixture 0.4 mole `H_2` and 0.2 mole `Br_2` is heated at 700 K at equilibrium, the value of equilibrium constant is `0.25xx10^10` then find out th

If a mixture 0.4 mole `H_2` and 0.2 mole `Br_2` is heated at 700 K at equilibrium, the value of equilibrium constant is `0.25xx10^10` then find out the ratio of concentrations of `(Br_2)` and (HBr) (Report your answer as `(Br_2)/(HBr)xx10^11`)

Monjur Alahi

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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 - 80
`{:(,H_2+," "Br_2" "hArr,2HBr),(t=0,0.4," "0.2,-),(t=t_(eq),0.2,underset("=negligible")(y),underset(=y)(0.4)):}`
`because 1/4xx10^10=(0.4xx0.4)/(0.2xxy)implies y=3.2xx10^(-10)implies (Br_2)/(HBr)xx10^11=3.2/0.2xx10^(-10)xx10^(11)=80`

4 views Answered 2023-02-05 15:29