Question

For the reaction:

Answer 1

Answer:

Half-life at 310 K is 6.54 × 10³ s.

Explanation:

Let's consider the following reaction:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

The rate law is:

(Δ[O₂]/Δt) = k . [N₂O₅]

Since [N₂O₅] is raised to the power of 1, the reaction order is 1.

For a first-order reaction:

$t_{1/2}=\frac{ln2}{k}$

where,

$t_{1/2}$ is the half-life

$k$ is the rate constant

At 300 K,

$(t_{1/2})_{1}=\frac{ln2}{k_{1}}\\k_{1}=\frac{ln2}{(t_{1/2})_{1}} =\frac{ln2}{2.50 \times 10^{4} s} =2.77 \times 10^{-5} s^{-1}$

We can use two-point Arrhenius equation to solve for k₂ at 310 K

$ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} (\frac{1}{T_{2}} -\frac{1}{T_{1}} )\\ln\frac{k_{2}}{k_{1}} =\frac{-103.3kJ/mol}{8.314 \times 10^{-3} kJ/mol.K} .(\frac{1}{310K}-\frac{1}{300K} )\\ln\frac{k_{2}}{k_{1}}=1.34\\k_{2}=1.06 \times 10^{-4} s^{-1}$

At 310 K,

$(t_{1/2})_{2}=\frac{ln2}{k_{2}}=\frac{ln2}{1.06 \times 10^{-4} s^{-1} } =6.54 \times 10^{3} s$