The equilibrium constant for the reaction N 2 g + O 2 g ⇌ 2 NO g At temperature T is 4 × 1 0 - 4 . the value of K C fro the reaction NO g ⇌ 1 2 N 2 g + 1 2 O 2 g at the same temperature is

N 2 g + O 2 g ⇌ 2 NO g K C = [ N O ] 2 [ N 2 ] [ O 2 ] = 4 × 1 0 - 4 NO ⇌ 1 2 N 2 g + 1 2 O 2 g K C = [ N 2 ] 1 / 2 [ O 2 ] 1 / 2 [ N O ] 1 K C = 1 × 1 0 4 4 = 50

Start Free Mock LoginAlready registered? Login here

Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

The equilibrium constant for the reaction
$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$
At temperature T is $4 \times 1 0^{- 4}$ . the value of $K_{C}$ fro the reaction
$NO (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ at the same temperature is

Options:

Answer:

Solution:

$N_{2} (g) + O_{2} (g) ⇌ 2 NO (g)$
$K_{C} = \frac{[N O]^{2}}{[N_{2}] [O_{2}]} = 4 \times 1 0^{- 4}$
$NO ⇌ \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$
$K_{C} = \frac{[N_{2}]^{1 / 2} [O_{2}]^{1 / 2}}{[N O]}$
$\sqrt{\frac{1}{K_{C}}} = \sqrt{\frac{1 \times 1 0^{4}}{4}} = 50$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Chemical EquilibriumSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...