The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours. g

2.858g

Explanation:

First order reaction.

Half life (t1/2) = 688

Specific rate constant, K = ?

The relationship between t1/2 and K id given as;

t1/2 = 0.693 / k

k = 0.693 / t1/2

k = 0.693 / 688 = 0.0010073 hr-1

Amount of 3.0 g sample of CS2 that remains after 48 hours.

Integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln[A] = ln[3] - 0.0010073(48)

ln[A] = 1.05026

[A] = 2.858g

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.

i'm wiling to try and you : )