Answer:
Second order
Explanation:
We could obtain the order of reaction by looking at the table very closely.
Now notice that in experiment 1 and 2, the concentration of [OH^-] was held constant while the concentration of [S8] was varied. So we have;
a situation in which the rate of reaction was tripled;
0.3/0.1 = 2.10/0.699
3^1 = 3^1
Therefore the order of reaction with respect to [S8] is 1.
For [OH^-], we have to look at experiment 2 and 3 where the concentration of [S8] was held constant;
x/0.01 = 4.19/2.10
x/0.01 = 2
x = 2 * 0.01
x = 0.02
So we have;
0.02/0.01 = 2^1
2^1 = 2^1
The order of reaction with respect to [OH^-] = 1
So we have the overall rate law as;
Rate = k[S8]^1 [OH^-] ^1
Overall order of reaction = 1 + 1 = 2
Therefore the reaction is second order.
