Initial rate of rxn of H2O2(aq)->H2O+ 1/2 O2(g) is found at 1.7x10(-3)M/s and is constant for 2min. You start with 160ml of 16M H2O2.
How many mL of O2 in one minute? (Measured at 273K and 760mmHg).
I tried to find the new conc of H2O2 via NewConc= Old conc - kt
Then i multiplied the new conc by .16L, converted those moles to O2 and solved for volume via the PV=nRT, but ended up with 2.3ishL, which seems rather large considering it's asking for V in mL. Any idea if this is right or not?
(I doubt i can assume the volume is constant in my method, among all the other things most likely wrong with it XD)
Thanks!
you did not balance the equation first. it is rather difficult to react 1/2 an atom.
No it's balanced. you're allowed to have fractions in the equation. Uncommon yes, but not impossible. Nice try though :P
To determine the amount of O2 produced in one minute, we need to find the change in concentration of H2O2 in one minute, and then use the stoichiometry of the reaction to calculate the corresponding change in the number of moles of O2.
Given:
Initial rate of reaction = 1.7 × 10^(-3) M/s
Time = 2 minutes
Initial volume of H2O2 = 160 mL
Initial concentration of H2O2 = 16 M
Final volume of H2O2 = 160 mL
First, let's find the change in concentration of H2O2 in two minutes:
Change in concentration of H2O2 = Initial rate × Time
Change in concentration of H2O2 = (1.7 × 10^(-3) M/s) × (2 min) = 3.4 × 10^(-3) M
Next, let's find the actual concentration of H2O2 after the reaction:
Final concentration of H2O2 = Initial concentration - Change in concentration
Final concentration of H2O2 = 16 M - 3.4 × 10^(-3) M
Now, let's convert the final concentration of H2O2 to moles:
Final moles of H2O2 = Final concentration × Final volume
Final moles of H2O2 = (16 M - 3.4 × 10^(-3) M) × (0.160 L)
To find the moles of O2 produced, we need to use the stoichiometry of the reaction, which states that one mole of H2O2 produces 1/2 mole of O2. Therefore:
Final moles of O2 = (1/2) × Final moles of H2O2
Finally, let's convert the moles of O2 to volume at the specified conditions (273 K and 760 mmHg) using the ideal gas law:
PV = nRT
V = (Final moles of O2 × R × T) / P
Plug in the values for R (0.0821 L·atm/mol·K), T (273 K), and P (760 mmHg) to calculate the volume of O2 in liters. Then convert it to milliliters.
It seems like you followed a similar approach, but keep in mind that the volume of the reaction may change due to the production of O2 gas. Therefore, it's necessary to account for the change in volume rather than assuming it remains constant.