Question

At a particular temperature the first-order gas-phase reaction N 2O 5→ 2NO 2 + (1/2)O 2 has a half-life for the disappearance of dinitrogen pentoxide of 5130 s. Suppose 0.450 atm of N 2O 5 is introducted into an evacuated 2.00 L flask. What will be the total gas pressure inside the flask after 3.00 hours? A. 0.969 atm B. 0.105 atm C. 0.795 atm D. 1.14 atm E. 0.864 atm

Answer 1

Answer:

B. 0.105 atm

Explanation:

Let's consider the following first-order reaction.

N₂O₅ → 2NO₂ + 1/2 O₂

We can find the pressure of N₂O₅ at a certain time t using the following expression.

$pN_{2}O_{5}=(pN_{2}O_{5})_{0}.e^{-k.t}$

where,

$(pN_{2}O_{5})_{0}$ is the initial pressure

k is the rate constant

Given the half-life (t1/2), we can calculate k using the following expression.

k = ln 2 / t1/2 = ln 2 / 5130 s = 1.351 × 10⁻⁴ s⁻¹

When t = 3.00 h (10,800 s),

$pN_{2}O_{5}=0.450atm.e^{-1.351 \times 10^{-4}s^{-1} \times 10,800 s } =0.105atm$