KC for N2+O2⇌ 2NO at certain temperature is 1.6×10−3. Then Kp for NO ⇌ N2+ O2 at same temperature will be ________

N2 + O2 ⇌ 2NO KC2NO ⇌ N2 + O2 1KC12 (2NO ⇌ N2 O2) ⇒ (1KC)1/2=1KC= 116 × 154 = 14 × 10-2 = 1004 = 25Kp = Kc (RT)Δ n NO ⇌ 12 N2 + 12O2Kp = Kc (RT)0 ∵Δn = (12 + 12) - 1KP = KC = 25 = 25

KC for N2+O2⇌ 2NO at certain temperature is 1.6×10−3. Then Kp for NO ⇌ N2+ O2 at same temperature will be ________

N2 + O2 ⇌ 2NO KC2NO ⇌ N2 + O2 1KC12 (2NO ⇌ N2 O2) ⇒ (1KC)1/2=1KC= 116 × 154 = 14 × 10-2 = 1004 = 25Kp = Kc (RT)Δ n NO ⇌ 12 N2 + 12O2Kp = Kc (RT)0 ∵Δn = (12 + 12) - 1KP = KC = 25 = 25

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K_Cfor $N_{2} + O_{2} \underset{}{⇌} 2 N O$ at certain temperature is $1.6 \times 10^{- 3} .$ Then K_p for $N O \underset{}{⇌} \frac{}{} N_{2} + \frac{}{} O_{2}$ at same temperature will be ________

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Detailed Solution

$N_{2} {+ O}_{2} \underset{}{⇌} {2NO K}_{C}$
$2NO \underset{}{⇌} N_{2} {+ O}_{2} \frac{1}{K_{C}}$
$\frac{1}{2} (2NO \underset{}{⇌} N_{2} O_{2}) \Rightarrow {(\frac{1}{K_{C}})}^{1 / 2} = \frac{1}{\sqrt{K_{C}}}$
$= \frac{1}{\sqrt{16 × 154}} = \frac{1}{{4 × 10}^{-2}} = \frac{100}{4} = 25$
$Kp = Kc {(RT)}^{Δ n} NO \underset{}{⇌} \frac{1}{2} N_{2} + \frac{1}{2} O_{2}$
$Kp = Kc {(RT)}^{0} ∵ Δn = (\frac{1}{2} + \frac{1}{2}) - 1$
$K_{P} {= K}_{C} = 25 = 25$