Answer:
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
