Decomposition Of Hydrogen Peroxide

By Author: Sherry Roberts
Total Articles: 77
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Figure 33a.3 shows a plot of temperature (k) against Time (s) in which the values have a linear relationship. The temperature rises steadily as the time passes by which show that the two parameters have a direct relationship. The slope of the linear graph is determined as 0.002232 K/s, the value of Y-intercept is 295.1 K, and the correlation factor is 0.9268 which shows a close relationship between the variables.
Figure 33a.2 shows a linear plot of Kobs with Klo where the two parameters have a direct relationship. An increase in KLo results to a corresponding rise in kobs. The slope of the graph is 0.01042, and the Y-intercept is -0.0002544. The correlation of the graph is 0.9877 which shows that the two compared variables have a close relationship and that a change in one results in a corresponding change in another.
Figure 33a.1 shows a plot of natural log of (Ln) term versus Time (s); in which there are three different lines of best fit for comparison of the change on the natural logarithm of term value as time changes. As evident from the plot, the natural log of term value decreases from zero to the negative ...
... side as the time increases. The plot shows that there two variables have an inverse relationship. The correlation values for the lines of best fit shows that the variables have a close relationship.
Table 33a.2 shows the experimental data obtained for three different runs at the same temperature of 294.5 but at varying concentrations of hydrogen peroxide. The other data values collected from the decomposition of hydrogen peroxide are [I-]0/ mol/ L, VL/ml, VG/ml, P0/kpa, and Ω/ kpa-1. The table also has the calculated slopes from the linear equations obtained from the experiment and the calculated kobs values.
Questions on Page 90 and 91 (1 to 6) Data Analysis
1. The kobs rate constant is as show in table 33a.2. for each of the three runs and are determined by linear regression.
2. The plot of pressure versus time should have curvature and equilibrate at a constant value since there is a point where a further increase in pressure does not result in any change in the other parameters, which is a point of equilibrium. In this case, the plot of the graph will have a curvature at a constant value. An increase in pressure on a reaction tends to increase the rate of reaction of the reacting gases. The increase in pressure of the gas is similar to increase its concentration. According to the Le Chatelier, an increase in pressure shifts the position of equilibrium that makes the pressure to reduce. The point of equilibrium is shown as a curvature in the plot of the graph.
3. Calculated values for ‘alpha’ as calculated using the below formula in each run is as shown below:
For Run 1, the value is 168/ 54758.799968 = 0.003068
For Run 2, the value is 128/ 61634.502475 = 0.002077
For Run 3, the value is 132/ 84271.43073 = 0.001566
4. The graph is as plotted in the attachment.
5. The slope must be divided by -2 to obtain the rate constant since its value has to be determined by linear regression which is in this has to be a positive value. Since the slope equals to -2kobs, it is important to eliminate the constant value of -2 from the value to obtain the rate constant by dividing the slope by -2.
6. The values of [KI] 0 are as calculated below for each run.
For Run 1, [KI] 0 equals to 0.00016
For Run 2, [KI] 0 equals to 0.00029
For Run 3, [KI] 0 equals to 0.000081

Sherry Roberts is the author of this paper. A senior editor at Melda Research in online nursing papers. If you need a similar paper you can place your order for a custom research paper from legitimate research paper writing service services.