For 2BrF5(g)⇌Br2(g)+5F2(g), Kp=1.024×10−14bar4 If at equilibrium, Br2=4×10−5bar and F2=4×10−4bar the partial pressure of BrF5 will be

BrF52=Br2F25Kp4×10−5 bar 4.0×10−4 bar 51.024×10−14 bar 4=4.096×10−25bar61.024×10−14bar4=4.0×10−8bar2BrF5=2.0×10−4bar

For 2BrF5(g)⇌Br2(g)+5F2(g), Kp=1.024×10−14bar4 If at equilibrium, Br2=4×10−5bar and F2=4×10−4bar the partial pressure of BrF5 will be

BrF52=Br2F25Kp4×10−5 bar 4.0×10−4 bar 51.024×10−14 bar 4=4.096×10−25bar61.024×10−14bar4=4.0×10−8bar2BrF5=2.0×10−4bar

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For $2 {BrF}_{5} (g) ⇌ {Br}_{2} (g) + 5 F_{2} (g), K_{p} = 1.024 \times 10^{- 14} {bar}^{4}$ If at equilibrium, $[{Br}_{2}] = 4 \times 10^{- 5} bar$ and $[F_{2}] = 4 \times 10^{- 4} bar$ the partial pressure of ${BrF}_{5}$ will be

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$2.0 \times 10^{- 4} bar$

$4.0 \times 10^{- 4} bar$

$4.0 \times 10^{- 8} bar$

$2.0 \times 10^{- 8} bar$

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${[{BrF}_{5}]}^{2} = \frac{[{Br}_{2}] {[F_{2}]}^{5}}{K_{p}} \frac{(4 \times 10^{- 5} bar) (4.0 \times 10^{- 4} {bar}^{5}}{(1.024 \times 10^{- 14} {bar}^{4})}$

$= \frac{4.096 \times 10^{- 25} {bar}^{6}}{1.024 \times 10^{- 14} {bar}^{4}} = 4.0 \times 10^{- 8} {bar}^{2}$

$[{BrF}_{5}] = 2.0 \times 10^{- 4} bar$

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For 2BrF5(g)⇌Br2(g)+5F2(g), Kp=1.024×10−14bar4 If at equilibrium, Br2=4×10−5bar and F2=4×10−4bar the partial pressure of BrF5 will be

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For $2 {BrF}_{5} (g) ⇌ {Br}_{2} (g) + 5 F_{2} (g), K_{p} = 1.024 \times 10^{- 14} {bar}^{4}$ If at equilibrium, $[{Br}_{2}] = 4 \times 10^{- 5} bar$ and $[F_{2}] = 4 \times 10^{- 4} bar$ the partial pressure of ${BrF}_{5}$ will be