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?
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
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
t = 107 minutes.
Therefore, time taken to decompose 95% of the CH3OCH3 is 107 minutes.