For the decomposition of hydrogen peroxide (H_2O_2) into water and oxygen gas, a
ID: 483816 • Letter: F
Question
For the decomposition of hydrogen peroxide (H_2O_2) into water and oxygen gas, an experiment yielded the following data: What is the order of the reaction with respect to H_2O_2? Explain the full reasoning behind your choice. Three graphs with your name on the print-out MUST be submitted in order to support your reasoning and earn credit for this problem. Determine the value of the rate constant for this reaction. Write the complete rate law for this reaction, including the value of k (with units): Calculate the half-life for this reaction. Does your answer make sense in light off the experimental data provided? Explain.Explanation / Answer
(a) At first, we do not know the order of the reaction.
So, we write the general expression for rate of a reaction
-dC/dt = kCn
Now, we consider the experimental data and substitute in the above equation
Case 1
-(0.693-0.879)/60 = k(0.879)n
3.1 * 10-3 = k(0.879)n
Case 2
-(0.378-0.458)/60 = k(0.458)n
1.33 * 10-3 = k(0.458)n
Dividing Case 1 by Case 2
(3.1 * 10-3)/(1.33 * 10-3) = (0.879/0.458)n
2.325 = (1.919)n
log(2.325) = n * log(1.919)
n = 1.29
(b) To find the rate constant
Taking Case 1
-(0.693-0.879)/60 = k(0.879)n
Now, substituting n = 1.29 , we get
k = 3.66 * 10-3
(c) Rate Law
-d[H2O2]/dt = r = k [H2O2]n
Substituting k and n,
-d[H2O2]/dt = 3.66 * 10-3 [H2O2]1.29
Units of k
We know, unit of reaction rate is mol/(liter.second)
and unit of concentration is mole/liter
So, r = kCn
mole/(liter.second) = k(mole/liter)1.29
Therefore, units of k is (mole/liter)-0.29 . (1/second)
(d) The half life of a reaction is defined as the time required for the concentration to reduce to half of its initial value.
Consider measurements 3 and 5
Concentration of measurement 3 (0.458) is almost double of that of 5 (0.236)
And, the time elapsed is (360 - 180) = 180 seconds
So, half life for the decomposition of hydrogen peroxide is 180 seconds.