For the reaction, $N_{2} O_{5} (g) \to N O_{2} (g) + \frac{1}{2} O_{2} (g)$ The value of rate of disappearance of $N_{2} O_{5}$ is given as $6.25 \times 10^{- 3} m o l L^{- 1} s^{- 1} .$ The rate of formation of $N O_{2}$ and $O_{2}$ is given respectively as:

Q. A mixture of

H_{2}, N_{2}

and

N H_{3}

with molar concentration

5 \times 10^{- 3} m o l L^{- 1}, 4 \times 10^{- 3} m o l L^{- 1} a n d 2 \times 10^{- 3} m o l L^{- 1}

respectively was prepared and heated to 500K.The value of

K_{c}

for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

at this given temperature is 60.Predict whether ammonia tends to form or decompose at this stage of concentration

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When 36.8g N2O4(g) is introduced into a 1.0-litre flask at 27∘C. The following equilibrium reaction occurs: N2O4(g)⇌2NO2(g);Kp=0.1642atm. Calculate Kc of the equilibrium reaction.

The correct option is A 6.667×10−3molL−1As we know,Kp=Kc(RT)ΔnKc=Kp(RT)−Δn=0.1642(0.083×300)−=6.667×10−3mol/L.

When 36.8g $N_{2} O_{4} (g)$ is introduced into a 1.0-litre flask at $27^{\circ} C$ . The following equilibrium reaction occurs: $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g); K_{p} = 0.1642 a t m$ . Calculate $K_{c}$ of the equilibrium reaction.

The correct option is A $6.667 \times 10^{- 3} m o l L^{- 1}$
As we know,
$K_{p} = K_{c} (R T)^{Δ n}$
$K_{c} = K_{p} (R T)^{- Δ n} = 0.1642 (0.083 \times 300)^{-} = 6.667 \times 10^{- 3} m o l / L .$