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Question: Nitrogen tetraoxide (N₂O₄) decomposes as: N₂O₄(g) → 2NO₂(g) If the pressure of N₂O₄ falls from 0.50...

Nitrogen tetraoxide (N₂O₄) decomposes as: N₂O₄(g) → 2NO₂(g)

If the pressure of N₂O₄ falls from 0.50 atm to 0.32 atm in 30 minutes, the rate of appearance of NO₂(g) is:

A

0.006 atm min⁻¹

B

0.003 atm min⁻¹

C

0.012 atm min⁻¹

D

0.024 atm min⁻¹

Answer

0.012 atm min⁻¹

Explanation

Solution

For the reaction

\ceN2O4(g)→2NO2(g)\ce{N_2O_4(g) \rightarrow 2NO_2(g)}

a drop in pressure of N₂O₄ is ΔP=0.50−0.32=0.18 atm\Delta P = 0.50 - 0.32 = 0.18\text{ atm} in 30 minutes. Thus, rate of disappearance of N₂O₄:

0.18 atm30 min=0.006 atm/min.\frac{0.18\text{ atm}}{30\text{ min}} = 0.006 \text{ atm/min}.

Since,

Rate=−11d[\ceN2O4]dt=12d[\ceNO2]dt,\text{Rate} = -\frac{1}{1}\frac{d[\ce{N_2O_4}]}{dt} = \frac{1}{2}\frac{d[\ce{NO_2}]}{dt},

we have the rate of appearance of NO₂:

d[\ceNO2]dt=2(0.006)=0.012 atm/min.\frac{d[\ce{NO_2}]}{dt} = 2(0.006) = 0.012\text{ atm/min}.