Dinitrogen pentoxide, N 2 O 5 , N2O5, decomposes by first‑order kinetics with a rate constant of 3.7 × 10 − 5 s − 1 3.7×10−5 s−1 at 298 K . 298 K. What is the half‑life, in hours, of N 2 O 5 N2O5 at 298 K ? 298 K? half‑life: h h If [ N 2 O 5 ] 0 = 0.0603 mol ⋅ L − 1 , [N2O5]0=0.0603 mol⋅L−1, what will be the concentration of N 2 O 5 N2O5 after 3.2 h ?

The half-life of the given reaction is 5.2 hours

The concentration of N₂O₅ after 3.2 hours is 3 × 10⁻² molL⁻¹

Explanation:

The value of rate constant, k is  3.7  ×  10⁻⁵  s⁻¹.

Temperature is  298  K  

The initial concentration of N₂O₅ is 6.03  × 10 ⁻²  m ol/L.

Time is 3.2 hours.

The formula for half-life of the first order reaction is given below:

t1/2 = 0.693 /k

Where,  t 1/2 is the half-life of first order reaction.

k  is the rate constant of the first order reaction.

Substitute the given values in the equation (I).

t1/2 =  0.693  / 3.7  ×  10⁻⁵  s⁻¹

t1/2 =  1.87  ×  10⁴ s

Converting to hours; (1.87  ×  10⁴ s  / 3600 s) * 1 hour = 5.2 hours

Therefore, the half-life of the given reaction is 5.2 hours

 The formula for first order reaction is shown as:

k  =  1 /t * ㏑([X]₀ / [X]ₙ)

Where,  k  is rate constant for first order reaction.

t  is time.

[X]₀ is initial concentration of the reactant.

[X]ₙ is concentration of the reactant at time t.

Converting 3.2 hours to seconds  3.2 * 60 * 60 = 11520 seconds

Substitute the given values in equation for the rate constant

3.7  ×  10⁻⁵  s⁻¹ = 1/11520 * ㏑(6.03  × 10 ⁻²  / [X]ₙ)

㏑(6.03  × 10 ⁻²  / [X]ₙ) =  3.7  ×  10⁻⁵  s⁻¹ * 11520

㏑(6.03  × 10 ⁻²  / [X]ₙ) = 0.426

6.03  × 10 ⁻²  / [X]ₙ = e ⁰°⁴²⁶

6.03  × 10 ⁻²  / [X]ₙ = 1.531

[X]ₙ = 6.03  × 10 ⁻² / 1.531

[X]ₙ = 3 × 10⁻² molL⁻¹

Therefore the concentration of N₂O₅ after 3.2 hours is 3 × 10⁻² molL⁻¹

c is the answer because when something is left after something evaporates that shows the chemical change.

the answer is 303.15 or 303