Given are the following informations
(i) ${BrF}_{3} (g) + F_{2} (g) ⇌ {BrF}_{5} (g) K_{p} = 8.6 \times 10^{35} {atm}^{- 1}$
(ii) $ClF (g) + F_{2} (g) ⇌ {ClF}_{3} (g) K_{p} = 7.8 \times 10^{12} {atm}^{- 1}$
(iii) $BrF (g) + F_{2} (g) ⇌ {BrF}_{3} (g) K_{p} = 7.8 \times 10^{27} {atm}^{- 1}$
The $K_{p}$ of the reaction $BrF (g) + 2 {ClF}_{3} (g) ⇌ 2 ClF (g) + {BrF}_{5} (g)$ will be

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$4.10 \times 10^{38}$

$1.10 \times 10^{38}$

$8.6 \times 10^{38}$

$5.52 \times 10^{16}$

answer is B.

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Detailed Solution

The given equation is obtained by the manipulationes Eq.(iii) + Eq.(i) - 2Eq(ii)

Hence $K_{p} = \frac{K_{p (iii)} K_{p (i)}}{K_{p (i i)}^{2}} = \frac{(7.8 \times 10^{27} {atm}^{- 1}) (8.6 \times 10^{35} {atm}^{- 1})}{{(7.8 \times 10^{12} {atm}^{- 1})}^{2}} = 1.10 \times 10^{38}$

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