# For the first order reaction CH3OCH3 ---> CH4 + CO + H2, it is found that a 150mg sample of CH3OCH3 is reduced to 43 mg in 45 minutes.How long would it take to decompose 95% of the CH3OCH3?

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### 1 Answer

CH3OCH3 ---->CH4 + CO + H2

We have been told that this is a first order reaction. That means the rate of reaction is directly proportionate to the reactant concentration.

r = k[CH3OCH3]

Therefore rate of reduction of CH3OCH3 is given by,

d[CH3OCH3]/dt = -k[CH3OCH3]

The given data is mass, but if we assume that the volume is constant then we can use mass in place of concentration. **So let m be the mass of CH3OCH3 at time t. Then,**

dm/dt = -km

separating variables,

dm/m = -kdt

**Integrating wrt t,**

ln(m) = -kt +c

To find c we can use the given data,

At t =0, m = 150 mg, then,

ln(150) = 0+c

**c = ln(150)**

ln(m) = -kt + ln(150)

ln(m/150) = -kt

We also that in 45 minutes, the sample was reduced to 43 mg.

ln(43/150) = -k x 45

**k = 0.028 per minute.**

Therefore the equation is given by,

**ln(m/150) = -0.028t**

If 95% of substance reacts, there is only 5% left.

Therefore mass left = 150 x 5 /100 = 7.5 mg

ln(7.5/150)= -0.028t

**t = 107 minutes.**

**Therefore, time taken to decompose 95% of the CH3OCH3 is 107 minutes.**

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