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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j