At equilibrium the concentration of nitrigen 3-0 - 10-3 molar- oxygen - 4-2 - 10-3 molar NO - 2-8 - 10-3 molar in a ceiled vesel at 800 k- What will be Kc for the reaction N2 - O2 - 2NO -

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Standard XII

Chemistry

Molality

At equilibriu...

Question

At equilibrium the concentration of nitrigen $3.0 \times 10^{- 3}$ molar, oxygen = $4.2 \times 10^{- 3}$ molar & $N O = 2.8 \times 10^{- 3}$ molar in a ceiled vesel at 800 k. What will be $K_{c}$ for the reaction $N_{2} + O_{2} \to 2 N O$ ?

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Solution

$N_{2} = 3.0 \times 10^{- 3}$ M
$O_{2} = 4.2 \times 10^{- 3}$ M
$N O = 2.8 \times 10^{- 3}$ M
Applying the law of chemical equilibrium we have
$K_{c} = \frac{[N O]^{2}}{[N_{2}] [O_{2}]}$ -----------(I)
Putting the values in repression (1), we have
$K_{c} = \frac{(2.8 \times 10^{- 3 M})^{2}}{(3.0 \times 10^{- 3} M) (4.2 \times 4.2 \times 10^{- 3} M)} = 0.622$

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At equilibrium the concentration of nitrigen 3.0×10−3 molar, oxygen = 4.2×10−3 molar & NO=2.8×10−3 molar in a ceiled vesel at 800 k. What will be Kc for the reaction N2+O2→2NO ?

N2=3.0×10−3MO2=4.2×10−3MNO=2.8×10−3MApplying the law of chemical equilibrium we haveKc=[NO]2[N2][O2]-----------(I)Putting the values in repression (1), we haveKc=(2.8×10−3M)2(3.0×10−3M)(4.2×4.2×10−3M)=0.622

At equilibrium the concentration of nitrigen $3.0 \times 10^{- 3}$ molar, oxygen = $4.2 \times 10^{- 3}$ molar & $N O = 2.8 \times 10^{- 3}$ molar in a ceiled vesel at 800 k. What will be $K_{c}$ for the reaction $N_{2} + O_{2} \to 2 N O$ ?