Question

The steam reforming of methane is an industrial which produces carbon monoxide and hydrogen as feedstocks for other materials. This reaction is normally run anywhere between 750 and 1100 C. When the reaction is run in a large excess of water, it displays first order kinetics with respect to the concentration of methane and zeroeth order kinetics with respect to water. At 750 degrees, the half-life of this reaction under these conditoins is 67.5 minutes. If the activitation energy for this reaction is 96.2 kj/mol: what is the rate contant, k, at both 750 and 1100 degrees celsius?

Answer 1

Solution :-

Overall reaction order is first order

So

At 750 C the half life is 67.5 minutes

Lets calculate the rate constant at the 750 C

Rate constant K = 0.693 / t1/2

                              = 0.693 / 67.5 min

                             = 0.01027 min-1

So the rate constant at the 750 C is   0.01027 min -1

Now lets calculate the rate constant at the 1100 C + 273 = 1373 K

T1 = 750 C + 273 = 1023 K

T2 = 1100 C + 273 = 1373 K

K1 = 0.01027 min-1

K2 = ?

Ea = 96.2 kJ /mol * 1000 J / 1 kJ = 96200 J per mol

Arrhenius equation

Ln[ K2/K1] = Ea/R [(1/T1)-(1/T2)]

Ln K2/0.01027 min-1] = 96200 J per mol / 8.314 J per mol K [(1/1023)-(1/1373)]

Ln K2/0.01027 min-1] = 2.883

K2/0.01027 min-1 = anti ln [2.883]

K2/0.01027 min-1 =17.86

K2 = 17.86 * 0.01027 min-1

K2 = 0.183 min-1

So the rate constant K at 1100 C is 0.183 min-1