C2H4O(g) �¨ CH4(g) + CO(g) Find the rate law of C2H4O?
The following kinetic data were observed for the reaction at 688 K.
Initial
Concentration
of Ethylene Oxide
Initial Rate
Exp. 1: 0.00272 M 5.57 10-7 M/s
Exp. 2: 0.00544 M 1.11 10-6 M/s
Here is how you do these.
rate = k(A)x where A in this case is concn ethylene oxide.
1.11E-6.....k*(0.00544)x
-------- = ----------------
5.57E-7.....k*(0.00272)x
k cancels
1.99 = (2)x
You may look at this and see that x must be 1 for the equality to hold OR you may solve it mathematically by taking the log of both sides.
log 1.99 = x*log 2
0.298 = 0.301x
x = 0.298/0.301 = 0.993 which I would round to 1. To determine k for the reaction, take EITHER trial run, plug in the values and evaluate k.
rate = k(A)1
Take run 2.
1.11E-6 = k(0.00544)1
and solve for k. The other run should give about the same k value.
To determine the rate law of the reaction C2H4O(g) → CH4(g) + CO(g), we need to analyze the relationship between the initial concentration of C2H4O and the initial rate of the reaction.
We are given two sets of kinetic data:
Exp. 1: Initial Concentration of C2H4O = 0.00272 M, Initial Rate = 5.57 × 10^(-7) M/s.
Exp. 2: Initial Concentration of C2H4O = 0.00544 M, Initial Rate = 1.11 × 10^(-6) M/s.
To determine the rate law, we need to compare the effect of changing the concentration of C2H4O while keeping the concentrations of other reactants constant (if there are any others) on the reaction rate.
Let's divide the rates of Exp. 2 by the rates of Exp. 1, while dividing the concentrations of C2H4O in Exp. 2 by the concentrations of C2H4O in Exp. 1:
Rate ratio = (1.11 × 10^(-6) M/s) / (5.57 × 10^(-7) M/s) = 2
Concentration ratio = (0.00544 M) / (0.00272 M) = 2
Since the concentration ratio is equal to the rate ratio, it suggests that the rate of the reaction is directly proportional to the concentration of C2H4O.
Therefore, the rate law for the reaction C2H4O(g) → CH4(g) + CO(g) is:
Rate = k[C2H4O], where k is the rate constant.