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 -

You visited us 12 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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 $N_{2} + O_{2} \to 2 N O$ ?

Open in App

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
$K_{c} = \frac{(2.8 \times 10^{- 3 M})^{2}}{(3.0 \times 10^{- 3} M) (4.2 \times 4.2 \times 10^{- 3} M)} = 0.622$

Suggest Corrections

Similar questions

Q. Given the reaction between

2

gases represented by

A_{2}

and

B_{2}

to give the compound

A B (g)

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

At equilibrium, the concentration of:

A_{2} = 3.0 \times 10^{- 3} M

B_{2} = 4.2 \times 10^{- 3} M

A B = 2.8 \times 10^{- 3} M

If the reaction takes place in a sealed vessel at

527^{\circ} C

, then the value of

K_{c}

will be:

Q. The following equilibrium concentrations were measured at

800 K : [S O_{2}] = 3.0 \times 10^{- 3} M; [O_{2}] = 3.5 \times 10^{- 3} M; [S O_{3}] = 5.0 \times 10^{3} M

Calculate the equilibrium constant at

800 K

for the chemical reaction in equilibrium for both directions.

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

(

K_{c}

is the equilibrium constant for the forward reaction and

K_{c}^{^{'}}

is the equilibrium constant for the backward reaction)

K_{C}

for

N_{2} + O_{2} ⇌ 2 N O

at certain temperature is

1.6 \times 10^{- 3}

, then

K_{P}

for

N O ⇌ 1 / 2 N_{2} + 1 / 2 O_{2}

at same temperature will be:

Q. The air pollutant NO is produced in automobile engines from the high temperature reaction

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); K_{c} = 1.7 \times 10^{- 3}

at 2300K. If the initial concentrations of

N_{2}

and

O_{2}

at 2300K are both 1.40M, what are the concentrations of

N O, N_{2},

and

O_{2}

when the reaction mixture reaches equilibrium?

Q. For the reaction

2 {N O}_{2} ⟶ N_{2} O_{2} + O_{2}

, rate expression is as follows;

- \frac{d [{N O}_{2}]}{d t} = k {[{N O}_{2}]}^{n}

, where

k = 3 \times 10^{- 3} {m o l}^{- 1} L {s e c}^{- 1}

. If the rate of formation of oxygen is

1.5 \times 10^{- 4} m o l L^{- 1} {s e c}^{- 1}

, then the molar concentration of

{N O}_{2}

in mole

L^{- 1}

is:

Join BYJU'S Learning Program

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 $N_{2} + O_{2} \to 2 N O$ ?