Question

The decomposition reaction $2N_2O_5(g) \xrightarrow{\Delta} 2N_2O_4(g) + O_2(g)$ is started in a closed cylinder under the isothermal isochoric condition at an initial pressure of 1 atm. After $Y \times 10^3$ s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is $5 \times 10^{-4} s^{-1}$, assuming ideal gas behaviour, the value of Y is

Answer 1

Y ≈ 4.61

Answer 2

Solution:

Let’s assume initially we have 2 moles of N₂O₅. The reaction is

$2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g)$.

If y moles of N₂O₅ have reacted, then by stoichiometry:

• Moles of N₂O₅ remaining = 2 – y.
• Moles of N₂O₄ formed = y [since 2 moles N₂O₅ produce 2 moles N₂O₄].
• Moles of O₂ formed = y/2 [since 2 moles N₂O₅ produce 1 mole O₂].

Thus, total moles at time t are: $n_{total} = (2 – y) + y + y/2 = 2 + y/2$.

Since the reaction is first order in N₂O₅, its decrease is given by: $[N_2O_5] = [N_2O_5]_0 e^{(–kt)}$

For our basis of 2 moles, $2 – y = 2 e^{(–kt)}$ ⇒ $y = 2(1 – e^{(–kt)})$.

Substitute in $n_{total}$: $n_{total} = 2 + [2(1 – e^{(–kt)})]/2 = 2 + (1 – e^{(–kt)}) = 3 – e^{(–kt)}$.

Under isothermal, constant volume conditions, pressure ∝ moles so: $P(t) = P_0 × (n_{total} / 2) = 1 atm × ((3 – e^{(–kt)}) / 2)$.

At time t = Y×10³ s, P(t) = 1.45 atm. Therefore, $(3 – e^{(–kt)}) / 2 = 1.45$ ⇒ $3 – e^{(–kt)} = 2.90$ ⇒ $e^{(–kt)} = 3 – 2.90 = 0.10$.

Taking logarithm: $–kt = ln(0.10)$ ⇒ $t = –ln(0.10)/k = (2.3026)/(5×10^{(–4)}) ≈ 4605 s$.

Since t = Y×10³ s, $Y ≈ 4605/1000 = 4.605 ≈ 4.61$.