Answered

Get detailed and reliable answers to your questions on IDNLearn.com. Our platform provides prompt, accurate answers from experts ready to assist you with any question you may have.

2. Methyl dichloroacetate (Cl2CHCO2CH3) decays into methanol (CH3OH) and dichloroacetic acid (Cl2CHCO2H) on reaction with water. Given a rate constant of 2.7 x 10-4/sec and an initial concentration of 1-ppm methyl dichloroacetate in the water, how much methanol will be present in the water after 30 min

Sagot :

Pstnonsonjokuidn

Answer:

2.69 * 10^-6 Mol/L

Explanation:

The equation of the reaction is;

Cl2CHCO2CH3 -------------------> Cl2CHCO2H + CH3OH

To convert from ppm concentration to Mol/L, we have

M = ppm/MM * 1000

Where;

M = mol/l

MM= Molar mass

M = 1/142.97 g/mol * 1000 =

M= 6.99 * 10^-6 Mol/L

For first order reaction;

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

Given that

[A]o = 6.99 * 10^-6 Mol/L

[A]=??

k= 2.7 x 10-4/sec

t= 30 mins * 60 = 1800 s

ln[A] = ln[6.99 * 10^-6] -   (2.7 x 10-4 * 1800)

ln[A] = -11.87 - 0.486

ln[A] = -12.356

[A] = e^(-12.356)

[A] = 4.3 * 10^-6 Mol/L

Concentration of methanol present after 30 mins= 6.99 * 10^-6 - 4.3 * 10^-6 = 2.69 * 10^-6 Mol/L

Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.