Determining the Rate Law for the Crystal Violet-Hydroxide ...



Determining the Rate Law for the Crystal Violet-Hydroxide Ion Reaction

Adam Capriola February 14, 2007

CHM 1122 Section 155

Introduction

The kinetics of a chemical equation is determined by its rate. The rate is the speed at which the reactants form into products. The rate is dependent on the concentrations and the orders of the reactants. One way to find the order is by first measuring the concentration of the products as time passes. A spectrophotometer is one tool that can measure relative concentration if the reactants change color as the form products. Graphing absorbance versus time, ln(absorbance) versus time, and 1/absorbance versus time will determine the order depending on which graph produces a straight line. In this experiment, crystal violet and NaOH form a complex that changes from transparent blue to colorless over time. The absorbance is measured using a spectrophotometer, and the rate law is then determined using this information.

Experimental

First, a spectrophotometer was turned on and set at a wavelength of 595 nm. Next, a cuvet was obtained, rinsed, and filled with deionized water. The outside of the cuvet was cleaned Kimwipe to get rid of smudges. The cuvet was then inserted into the spectrophotometer and the spectrophotometer was zeroed. Next, 10.0 mL of 0.010 M NaOH solution was dispensed into one clean 25 mL graduated cylinder and 10.0 mL of 1.50 x 10-5 M crystal violet solution was dispensed into another clean 25 mL graduated cylinder. The solutions were then simultaneously poured into a clean 50 mL beaker. This mixture was mixed with a glass stirring rod for a few moments to ensure consistency. The cuvet was then rinsed with the mixture two or three times and was then filled with the mixture. The cuvet again cleaned with a Kimwipe and was inserted into the spectrophotometer. The absorbance reading was measured every minute for twenty minutes, starting when the cuvet was first put in. This process was then repeated, replacing the 0.010 M NaOH solution with 0.020 M NaOH solution.

Results

0.010 M NaOH Solution:

|Time |Absorbance |lnA |1/A |

|0 |0.743 |-0.297 |1.35 |

|1 |0.728 |-0.317 |1.37 |

|2 |0.713 |-0.338 |1.40 |

|3 |0.703 |-0.352 |1.42 |

|4 |0.691 |-0.370 |1.45 |

|5 |0.680 |-0.386 |1.47 |

|6 |0.673 |-0.396 |1.49 |

|7 |0.664 |-0.409 |1.51 |

|8 |0.654 |-0.425 |1.53 |

|9 |0.644 |-0.440 |1.55 |

|10 |0.636 |-0.453 |1.57 |

|11 |0.624 |-0.472 |1.60 |

|12 |0.610 |-0.494 |1.64 |

|13 |0.603 |-0.506 |1.66 |

|14 |0.593 |-0.523 |1.69 |

|15 |0.580 |-0.544 |1.72 |

|16 |0.571 |-0.560 |1.75 |

|17 |0.560 |-0.580 |1.79 |

|18 |0.548 |-0.601 |1.82 |

|19 |0.539 |-0.618 |1.86 |

|20 |0.530 |-0.635 |1.89 |

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0.020 M NaOH Solution:

|Time |Absorbance |lnA |1/A |

|0 |0.697 |-0.361 |1.43 |

|1 |0.650 |-0.431 |1.54 |

|2 |0.606 |-0.501 |1.65 |

|3 |0.569 |-0.564 |1.76 |

|4 |0.530 |-0.635 |1.89 |

|5 |0.492 |-0.709 |2.03 |

|6 |0.459 |-0.779 |2.18 |

|7 |0.428 |-0.849 |2.34 |

|8 |0.399 |-0.919 |2.51 |

|9 |0.373 |-0.986 |2.68 |

|10 |0.351 |-1.047 |2.85 |

|11 |0.328 |-1.115 |3.05 |

|12 |0.308 |-1.178 |3.25 |

|13 |0.288 |-1.245 |3.47 |

|14 |0.269 |-1.313 |3.72 |

|15 |0.254 |-1.370 |3.94 |

|16 |0.236 |-1.444 |4.24 |

|17 |0.221 |-1.510 |4.52 |

|18 |0.206 |-1.580 |4.85 |

|19 |0.192 |-1.650 |5.21 |

|20 |0.179 |-1.720 |5.59 |

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Discussion/Conclusions

For the graphs using 0.010 M NaOH, the plot of Absorbance vs. Time had the straightest line (R2 = 0.9985), but the plot of lnA vs. Time also had a very straight line (R2 = 0.9971). The plot of 1/A had the least straight line with R2 equaling 0.9906. For the graphs using 0.020 M NaOH, lnA vs. Time had the straightest line (R2 = 0.9998). Absorbance vs. Time and 1/A vs. Time were not nearly as straight, with their R2 equaling 0.9646 and 0.9695, respectively. I think it is fairly safe to say that the rate equation is first order because lnA vs. Time overall yielded the straightest line in the two runs. The rate equation is rate = k [NaOH].

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