You visited us 2 times! Enjoying our articles? Unlock Full Access!

Characteristics of Equilibrium Constant

Calculate the...

Question

Calculate the value of the equilibrium constant for the reaction below if there are present at equilibrium 5.0 mol of $N_{2}$ , 7.0 mol of $O_{2}$ , and 0.10 mol of $N O_{2}$ in a 1.5 L vessel at a certain temperature.
$N_{2} + 2 O_{2} \to 2 N O_{2} + h e a t$

Open in App

Solution

$N_{2} (g) + O_{2} (g) ⟶ 2 N O (g) + h e a t$
Initially $5$ $7$ $0$
at equilibrium $5 - 0.1$ $7 - 0.1$ $0.1$
So,
$K_{C} = \frac{{[N O]}^{2}}{[N_{2}] {[O_{2}]}^{2}}$
$\Rightarrow K_{C} = \frac{{[0.1]}^{2}}{(5 - 0.1) {(7 - 0.1)}^{2}}$
$\Rightarrow K_{C} = 4.28$ mol/L
Total volume is $1.5 L$ .

Suggest Corrections

Similar questions

N_{2} + 2 O_{2} ⇌ 2 {N O}_{2}

100

2 {N O}_{2} ⇌ N_{2} + 2 O_{2} . . . . (i)

{N O}_{2} ⇌ 1 / 2 N_{2} + O_{2} . . . (i i)

K_{c} = 4 \times 10^{- 2}

N_{2} (g) + O_{2} (g) ⇌ 2 N O (g)

1.5 m o l

2 m o l

N_{2} a n d O_{2}

5 L

N O

m o l / L

N_{2} + 2 O_{2} ⇌ 2 N O_{2}

100

2 N O_{2} ⇌ N_{2} + 2 O_{2}

K_{1}

N O_{2} ⇌ \frac{1}{2} N_{2} + O_{2}

K_{2}

K_{1} \times K_{2}

N_{2}

O_{2}

N_{2} O

K_{C} = 2.0 \times 10^{- 37}

N_{2} + O_{2} ⇌ 2 N O

N O ⇌ \frac{1}{2} N_{2} + \frac{1}{2} O_{2}

Join BYJU'S Learning Program