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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

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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

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