Question

In order to determine the rate law for the decomposition of hydrogen peroxide, a student collected pressure versus time measurements as the hydrogen peroxide decomposed with the addition of Kl. The following graph was produced at 2.9 °C. What is the rate of formation, in M/s, of Oz? y = 3.2 x+7664.7 R2 = 0.9631 Pressure versus Time 850 840 830 820 810 Pressure (tor) 800 790 780 770 760 0 20 40 80 100 120 60 times

Answer 1

The rate of formation of Oz is 3.2 M/s.

Step-by-step explanation:

The given data provides a linear equation y = 3.2x + 7664.7 (R² = 0.9631), representing the pressure versus time measurements during the decomposition of hydrogen peroxide with the addition of Kl. The coefficient '3.2' in the equation corresponds to the rate of formation of Oz in M/s. This coefficient represents the slope of the line, indicating the change in pressure (which is directly proportional to the concentration) of Oz per unit time. Therefore, the rate of formation of Oz is 3.2 M/s.

The pressure versus time measurements demonstrate a linear relationship between the two variables, with a slope of 3.2. In kinetics, the rate of formation of a product in a chemical reaction is directly related to its concentration change over time. Here, the coefficient '3.2' derived from the linear equation signifies that the concentration of Oz increases by 3.2 M per second as time progresses during the hydrogen peroxide decomposition.

This value represents the rate constant for the formation of Oz and is indicative of a first-order reaction with respect to the concentration of hydrogen peroxide. The coefficient extracted from the linear equation directly gives the rate of formation of Oz in terms of M/s, hence yielding a rate of 3.2 M/s.