Question

The thermal decomposition of dimethyl ether has been studied by measuring the increase of pressure with time

Me2O = CH4 + H2+ CO

The following data were obtained

Time/sec 390 780 1195 2000 3155
Pressure increase/Torr 96 179 250 363 467

Show that the reaction is first order and find the rate constant (hint: Include P0 as a pramater as it is unknown)

Answer 1

Me2O = CH4 + H2+ CO

Let the initial pressure of dimethyl ether be Po.

The concentration of dimethyl ether is:

n / V = P_E / RT

where P_E the partial pressure of dimethyl ether.

At any stage in the reaction,

P(total) = P_E + P_M + P_H + P_C

P_E , P_M , P_H & P_C are the partial pressures of  ether, methane, hydrogen and CO, respectively.

1 mole dimethyl ether gives 1 mole methane, hydrogen and CO each.

If we start with pure ether, then

P_H = P_M = P_C

So, Po - PE = P_M = P_H = P_C.

P(total) = P_E + 3 (Po - P_E)

= 3Po - 2P_E

P_E = 1/2 * (3Po - P_tot)

= 1.5 Po - 0.5 * P_tot

P = P_tot - Po

P_E = Po - 0.5 P

Plotting  ln P_E vs t, we get:

Since, the graph of ln P_E vs t is a straight line.

So, the reaction is first order.

The slope of the line is given as:

- k = - 4.39x10^-4

Hence, the rate constant is: k = 4.39x10^-4 sec^-1.