K_C for $N_{2} + O_{2} ⇋ 2 N O$ at certain temperature is 1.6 × 10^-3, then K_P for $N O ⇋ 1 / 2 N_{2} + 1 / 2 O_{2}$ at same temperature will be.

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$25 atm$

$5 atm$

$1.6 \times 10^{- 3} {atm}^{- 1}$

answer is A.

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Detailed Solution

Given: it is given that $K_{c} = 1.6 \times 10^{- 3} a t m$ for reaction 1st: $N_{2} + O_{2} ⇌ 2 N O$

We have to find $K_{p}$ for reaction 2nd: $N O ⇌ 1 / 2 N_{2} + 1 / 2 O_{2}$

Formula Used: $K_{p} = K_{c} (R T)^{Δ n}; Δ n = n_{p} - n_{r}$

Explanation: For reaction 1st, $n_{p} = 2$ and $n_{r} = 1 + 1 = 2$ ; therefore, $Δ n = 2 - 2 = 0$ and $K_{p} = K_{c} = 1.6 \times 10^{- 3}$

In reaction 2nd, NO gives $1 / 2 N_{2} + 1 / 2 O_{2}$ i.e., ½ of the inverse of reaction 1st,

Therefore, $K_{p}' = 1 / (K_{p})^{1 / 2} = 1 / (1.6 \times 10^{- 3})^{1 / 2} = 25$

Hence, the correct option is A i.e., 25.

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