3S8 + 8 OH- + 8 S3 + 4 HOOH In an experiment, a student studies the kinetics of the reaction represented above and
obtains the data shown in the following table.
Initial [S
Initial
Initial (OH)
Experiment
Reaction Rate
(M)
(M)
(M/s)
Trial 1
0.100
0.0100
0.699
Trial 2
0.300
0.0100
2.10
Trial 3
0.300
?
4.19
Use the data in the table and determine the order of the reaction with respect to Sg.
O first order
O zero order
O second order
O third order

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.

radioactive decay occurs in unstable atomic nuclei – that is, ones that don't have enough binding energy to hold

the nucleus together due to an excess of either protons or neutrons.

answer: i dont know the answer but this might

explanation:

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