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 transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

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