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?

half life =   , are taken for the concentration to decrease to 25% and for the concentration to decrease to 6.25% .

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

half life =

(where k stands for rate constant)

Value of k is given as

half life =

half life = 1117.74 min or

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:

Where, 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.

t = 2241.94 min or

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

t = 4483.87 min or

So, the answers for all the three parts are: half life =   , are taken for the concentration to decrease to 25% and for the concentration to decrease to 6.25% .

the product of the pressure and volume is also constant

14 days is the half-life of phosphorus-32