Answer to Question #168112 in Chemistry for bella

Question

a reaction is first order in hydrogen peroxide, H2O2. It is determined that it takes 2.00 minutes for the [H2O2] to go from 6.0 M to 0.625 M at a certain temperature. What is the half-life of H2O2? What is the specific rate constant?

Answer 1

[H₂O₂]_o = 6.0 M

[H₂O₂] = 0.625 M

t = 2.00 min

Solution:

Decomposition of hydrogen peroxide (H₂O₂) is 1^st order reasction.

Integrated form of first-order rate law:

ln(A) = -kt + ln(A_o)

Thus,

ln[H₂O₂] = -kt + ln[H₂O₂]_o

ln(0.625) = -k × 2.00 + ln(6.0)

-0.47 = -k × 2.00 + 1.79176

k = 1.13088 min^-1

Half-life period of a first order reaction is given by t_1/2 = 0.693 / k.

Hence,

t_1/2 = 0.693 / 1.13088 = 0.6128

t_1/2 = 0.6128 min = 36.77 sec

Answer:

The half-life of H₂O₂ is 0.6128 min (or 36.77 sec);

The specific rate constant (k) is 1.13088 min^-1.