Question

If the rate constant for the decomposition of N2O5 is 6.2 × 10−4/min, what is the half-life? (The rate law is first order in N2O5.) How long would it take for the concen- tration of N2O5 to decrease to 25% of its initial value? to 6.25% of its initial value?

Answer 1

Answer: half life =  $1.1*10^3min$ , $2.24*10^3min$ are taken for the concentration to decrease to 25% and $4.48*10^3min$ for the concentration to decrease to 6.25% .

Explanation: The given information says, the reaction is first order with respect to $N_2O_5$ . For first order reaction, rate constant and half life are related to each other as:

half life = $\frac{0.693}{k}$

(where k stands for rate constant)

Value of k is given as $6.2*10^-^4min^-^1$

half life = $\frac{0.693}{6.2*10^-^4min^-^1}$

half life = 1117.74 min or $1.1*10^3min$

Let's say the initial concentration is 100. It asks to calculate the time taken to decrease the concentration to 25% which will be 25 as we have taken the initial concentration 100.

The equation that we use is:

$lnA=lnA_0-kt$

Where, $A_0$ is initial and A is remaining concentration or amounts. k is rate constant and t is the time. Let's plug in the values and do calculations for t.

$ln25=ln100-6.2*10^-^4min^-^1*t$

$3.22=4.61-6.2*10^-^4min^-^1*t$

$3.22-4.61=-6.2*10^-^4min^-^1*t$

$-1.39=-6.2*10^-^4min^-^1*t$

$t=\frac{1.39}{6.2*10*^-^4min^-^1}$

t = 2241.94 min or $2.24*10^3min$

We can calculate the time when the concentration decreases to 6.25% of its initial value same as we did for the above.

$ln6.25=ln100-6.2*10^-^4min^-^1*t$

$1.83=4.61-6.2*10^-^4min^-^1*t$

$1.83-4.61=-6.2*10^-^4min^-^1*t$

$-2.78=-6.2*10^-^4min^-^1*t$

$t=\frac{2.78}{6.2*10*^-^4min^-^1}$

t = 4483.87 min or $4.48*10^3min$

So, the answers for all the three parts are: half life =  $1.1*10^3min$ , $2.24*10^3min$ are taken for the concentration to decrease to 25% and $4.48*10^3min$ for the concentration to decrease to 6.25% .