Question

A sample of 0.900 mol N2O is placed in a sealed
container, where it decomposes irreversibly to N2 and O2
in a first-order reaction. After 42.0 min, 0.640 mol N2O
remains. How long will it take for the reaction to be
90.0% complete?

Answer 1

Answer:

• t = 284 min (rounded to 3 significant figures).

Explanation:

1. Decomposition reaction:

Chemical equation:

• 2N₂O → 2N₂ + O₂

2. First order reaction:

Rate law:

$r=-\frac{d[N_2O]}{dt}= k[N_2O]\\ \\ \frac{d[N_2O]}{dt}= -k[N_2O]$

Integrated rate law:

$\frac{d[N_2O]}{[N_2O]}=-kdt\\ \\ ln[N_2O]-ln[N_2O]_0=-kt$

3. Determine the rate constant, k

• t = 42.0 min, [N₂O] = 0.640 mol

• [N₂O]₀ = 0.900

• $ln(0.640)-ln(0.900)=-k(42.0min)$

• k = 0.00812 min⁻¹

4. Time taken for the reaction to be 90.0% complete

• [N₂O] = 0.100[N₂O]=0.100×0.900 mol = 0.0900 mol

• ln(0.0900) - ln(0.900) = - 0.00812 min⁻¹ × t

• t = (2.30 / 0.00812) min ≈ 283.6 min ≈ 284 min (rounded to 3 significant figures).

Answer 2

The time taken to have 90% of the reaction complete is 16429 s or 274 mins.

Using the relation;

N(t) = Noe^-kt

N(t) = amount at time t

No = amount initially present

k = decay constant

t  =  time taken

Now;

0.640 =  0.900e^-2520k

0.640/ 0.900 = e^-2520k

ln(0.640/ 0.900) = ln(e^-2520k)

ln(0.71) = -2520k

k = ln(0.71)/-2520

k = 1.4 * 10^-4 s-1

Hence at N(t) we have No - 0.9No = 0.1No

0.1No = Noe^-1.4 * 10^-4t

0.1No/No = e^-1.4 * 10^-4t

0.1 = e^-1.4 * 10^-4t

ln(0.1) = lne^-1.4 * 10^-4t

-2.3 = -1.4 * 10^-4t

t = -2.3/-1.4 * 10^-4

t = 16429 s or 274 mins

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