For the reaction:
2n2o5(g) → 4no2(g) + o2(g) the rate law is: (δ[o2]/δt) = k[n2o5] at 300 k, the half-life is 2.50 × 104 seconds and the activation energy is 103.3 kj/mol. what is the half-life at 310 k? (hint: use rate law expression to determine the reaction order → solve for k1 at 300 k using the corresponding half-life expression → use two-point arrhenius equation to solve for k2 at 310 k → use the half-life expression again to solve for half-life at 310 k)

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:

where,

is the half-life

is the rate constant

At 300 K,

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

At 310 K,

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a