Can you help with the question marks below and explain how to get the answer ple
ID: 883007 • Letter: C
Question
Can you help with the question marks below and explain how to get the answer please!!!
Table 1: 10 mL Undiluted IKI and 5 mL 3% H2O2
Slope:15.073
Inverse Slope:0.066
Table 2: 10 mL 0.5–1.0% IKI and 5 mL 3% H2O2
Slope: 23.062
Inverse Slope: 0.043
Table 3:10 mL 0.5–1.0% IKI and 5 mL 2.25% H2O2
Slope: 22.599
Inverse Slope: 0.044
Order of H2O2: ??????????????????
Order of IKI:????????????????????
Rate law constant: ????????????????
Table 1. 10 mL Undiluted IKI and 5 mL 3% H2O2 Water displaced in milliliters Time in seconds 4 45.4 6 64.5 8 98.1 10 122.6 12 138.1 14 150.5 16 196.7 18 241.8 20 277 22 326.9 Table 2. 10 mL 0.5–1.0% IKI and 5 mL 3% H2O2 Water displaced in milliliters Time in seconds 2 88.6 4 129.5 6 149.3 8 197.3 10 236.1 12 268.5 14 331.6 16 393.7 18 448.9 20 499.5 22 529.9 Table 3. 10 mL 0.5–1.0% IKI and 5 mL 2.25% H2O2 Water displaced in milliters Time in seconds 2 85.8 4 129 6 147.6 8 196.8 10 219 12 246 14 327.6 16 391.8 18 448 20 493 22 509Explanation / Answer
The slope given in all three cases are +ve, so the reaction is a second order reaction. Since for zero order and first order the slope is always -ve.
for a second order reaction,
1/[A] = 1/[Ao] + kt
or, y = mx + b
[A] = concentration at time t
[Ao] = initial concentration
k = rate constant = slope
t = time
From Table 1 and Table 2 data, when the concentration of IKI is changed the slope also changed (inverse relation as seen from the equation). From Table 2 and Table 3 data, where the concentration of H2O2 is altered, the slope almost did not change. So the reaction rate is dependent only on the concentration of IKI.
Order of H2O2 = 0
Order of IKI = 2
Rate law = k[IKI]^2