Question

FC1O₂(g) → FC10(g) + O(g)

The first-order decomposition of FC1O₂ (g) is represented by the equation above. At a certain temperature, the partial pressure of FC1O2(g) in a sealed vessel falls from 0.080 atm to 0.010 atm over 48
minutes.
What is the half-life of the decomposition reaction?

Answer 1

The half-life of a first-order reaction is given by the following equation:

t(1/2) = ln(2)/k

where t(1/2) is the half-life and k is the rate constant.

From the given information, we know that the partial pressure of FC1O2(g) falls from 0.080 atm to 0.010 atm over 48 minutes. We can use this information to calculate the rate constant as follows:

ln([FC1O2]t/[FC1O2]0) = -kt

where [FC1O2]t is the concentration (in atm) at time t, [FC1O2]0 is the initial concentration (in atm), and k is the rate constant.

Substituting the given values, we get:

ln(0.010/0.080) = -k(48)

Solving for k, we get:

k = 0.1225 min^-1

Now, we can use the equation for the half-life to calculate the answer:

t(1/2) = ln(2)/k

t(1/2) = ln(2)/0.1225

t(1/2) = 5.66 min (rounded to two significant figures)

Therefore, the half-life of the decomposition reaction is 5.66 minutes.