Question

q2. manganese dioxide catalyses the decomposition of hydrogen peroxide solution. oxygen and water are produced. (a) explain how a manganese dioxide catalyst increases the rate of decomposition of hydrogen peroxide. (b) add 50 cm³ of 2.0 mol/dm³ hydrogen peroxide solution to a conical flask. 2. add 1.0 g of manganese dioxide to the conical flask. 3. place the conical flask on a balance and start a timer. 4. record the total mass lost from the conical flask every 20 seconds for 180 seconds. (c) figure 1 shows the results for 50 cm³ of 2.0 mol/dm³ hydrogen peroxide solution and 1.0 g of manganese dioxide. a tangent to the line has been drawn at 75 seconds. determine the rate of reaction when the time was 75 seconds. give your answer to 2 significant figures. (d) explain why the mass of the conical flask and contents decreased. (e) fine manganese dioxide powder produces a higher rate of reaction than coarse manganese dioxide lumps. sketch on the figure above the results you would expect for student bs experiment with fine manganese dioxide powder. hydrogen peroxide molecules collide with manganese dioxide particles during the reaction. (f) why does fine manganese dioxide powder produce a higher rate of reaction than coarse manganese dioxide lumps? tick (✓) one box. fine manganese dioxide powder has a larger surface area. fine manganese dioxide powder has larger particles. fine manganese dioxide powder produces less frequent collisions. (total 10 marks)

Explanation:

Step1: Understand the reaction concept

Manganese dioxide catalyzes hydrogen - peroxide decomposition. A catalyst provides an alternative reaction pathway with lower activation energy.

Step2: Analyze surface - area effect

Fine manganese dioxide powder has a larger surface area. Larger surface area means more sites for reactant molecules (hydrogen - peroxide) to attach and react, increasing the frequency of collisions and thus the reaction rate.

Step3: Determine rate from graph

To find the rate of reaction at 75 seconds from the graph, we consider the slope of the tangent at that point. The slope of the tangent to the mass - time graph gives the rate of change of mass (which is related to the rate of reaction as mass loss is due to oxygen evolution). If we assume the general formula for the rate of reaction from a mass - time graph: Rate = $\frac{\Delta m}{\Delta t}$, where $\Delta m$ is the change in mass and $\Delta t$ is the change in time. From the graph, we estimate the change in mass over a small time interval around 75 seconds for the tangent line. Let's say we estimate the change in mass $\Delta m$ over a $\Delta t = 25$ - second interval (for example, from 62.5 s to 87.5 s) corresponding to the tangent. If the mass changes by approximately 0.2 g in this 25 - second interval, then Rate = $\frac{0.2}{25}= 0.008$ g/s. But we need to convert to cm³/s. Since the mass loss is due to oxygen evolution and we know the density of oxygen (at standard conditions, density of $O_2\approx1.43$ g/L or 0.00143 g/cm³), we first find the volume of oxygen evolved. If mass of oxygen = 0.2 g, then volume of oxygen $V=\frac{m}{
ho}=\frac{0.2}{0.00143}\approx140$ cm³. So the rate in cm³/s is $\frac{140}{25}=5.6$ cm³/s. Rounding to 2 significant figures, the rate is 5.6 cm³/s.

Answer:

5.6 cm³/s