Methane gas reacts with diatomic sulfur in the gas phase, producing carbon disulphide and hydrogen sulfide. The rate constant for this reaction is 1.1 (L/M.s) at 550 oC. If it is 6.31 (L/M.s) at 624 oC, what is the value of the activation energy for this reaction.
You can't plug all of this into the Arrhenius equation?
To determine the activation energy for this reaction, we can use the Arrhenius equation:
k = A * exp(-Ea / (RT))
Where:
- k is the rate constant,
- A is the pre-exponential factor or the frequency factor,
- Ea is the activation energy,
- R is the gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin.
We have two sets of data for the rate constant (k) at two different temperatures (550 oC and 624 oC). We can set up two equations using the Arrhenius equation:
k1 = A * exp(-Ea / (R * T1))
k2 = A * exp(-Ea / (R * T2))
Since the Arrhenius equation contains an exponential term, we can take the natural logarithm (ln) of both sides to simplify the equation:
ln(k1) = ln(A) - (Ea / (R * T1))
ln(k2) = ln(A) - (Ea / (R * T2))
Now we can subtract these two equations to eliminate the pre-exponential factor A:
ln(k1) - ln(k2) = -Ea / (R * T1) + Ea / (R * T2)
Rearranging the equation gives us:
ln(k1 / k2) = Ea / R * (T2 - T1) / (T1 * T2)
Now we can plug in the values we have:
ln(1.1 / 6.31) = Ea / (8.314 J/(mol·K)) * ((624 + 273) - (550 + 273)) / ((550 + 273) * (624 + 273))
Calculating the right side of the equation gives us:
ln(1.1 / 6.31) = Ea / (8.314 J/(mol·K)) * (897 - 823) / (823 * 897)
ln(1.1 / 6.31) = (Ea / (8.314 J/(mol·K))) * 0.096
Now we can isolate Ea and solve for it. We rearrange the equation once more:
Ea = (8.314 J/(mol·K)) * ln(1.1 / 6.31) / 0.096
Ea ≈ 208.8 J/mol
Therefore, the approximate activation energy for this reaction is 208.8 J/mol.