Question

my chemistry lab did the composition of hydrogen peroxide. i am having trouble trying to fugure out how to fill in the table. i dont understand why “s” is added to pvnrt and how to apply that to my equation. these are the steps on how to solve but i believe they are rather confusing. please help! thank you!

i dont know how to convert PV=nRT so m/s for the initial rate. im confused on how to fill out the swcodn table. hopefully this helps

Data anu Dservaions De motuor umtS ano srg Initial rate Total Run Reactants Temperature (kPa/s) volume ml 3.0 % H,0 , + mL 0.5 M KI 2 mL H,O mL. 3.0 % H,O , +S mL 0.5 M KI + 294co.0.016 has 2mL 3.0% H,O, .m. 0.5 M Kl+m H,O094A G0.0 054 hlalod CO 2nnulo horbo + 95ml H0262160lo. 0976 hPawb helom olao 4 mL 3.0 % H,0, . mL 0.5 M KI + mL p6 capcarsrGa o1 Analysis Initial rate Run (M/s) after mixing after mixing I Rate constant k H2O2 1 2 3 ai ti ADVICE FOR FILLING IN THE TABLE. 1) Rates of all chemical processes are conventionally kPa/s unit Therefore, you want to convert the 4 Recall that the ideal gas law is: PiCuehmedume aqual moesin time pcan Pu-naT CS CTanperas Divide both perts of the ideal gas law by V*R*TS What does molarity (moles over iters) per 2) second equal to, based on your rearranged equation? M/3= VA Use vour fnding above to calculate the first column of the analysis table, Use the iterature vale 3) of 8314 PaKmol for the ideal gas constant R. Show here how to caloulate the nun 1's veae 0.014 Afas omL) O:1i9 4(2a43) n(.134LPa/ mats n( 2397,9032 4) The moment you mixed your 3% (by weight) H,O, with 0.cc024 moes other solutions, its concentration (underline) decreased/increased. You can always use calculate molarity changes during dilutions. As long as you use same units for volume on both sides, formula will work. 3% (by weight) HO, is the same as 0.88 M. Show here how to calculate the run 1's [HO] value. M- Goc029 mdl O.050L M=0.0059ao M,V-M,V, to mol are same values, because KI is a soluble salt, 5) [KI] and [I] strong electrolyte. In water, it will dissociate completely into K' and I. Therefore, [KI]-[K'] = [I]. Show here how to calculate the run 1's [I] value. 6) For calculation of k's value (rate constant), assume it's a first order reaction by each of the reagents. That is, rate k[H.O.][I].Based on this equation, how can k be calculated and what are its units? Show here how to calculate the run l's k value. 7) Does k value vary for runs 1-32 Comment if it's in line our out of line with your expectations. Data and Observations (Be mindful of units and significant figures) Initial rate Total volume Temperature Reactants Run (kPa/s) 1 50 mL H,O mL 0.5 M KI+ ,mL 3.0% H,O, +S 2 OmL 3.0% H,O, mL H2O mL 0.5 M KI + + 10 mL 0.5 M KI+mL H,O mL 3.0% H2O2 + 4 GO mL 3.0% H2O2 mL 0.5 M KI+ IGML + ctom bousluoles oddnsp Analysis Initial rate [H2O,] ] Rate constant Ru(M/s) after mixing after mixing 1 2 o onit i a'i li ADVICE FOR FILLING IN THE TABLE. 1) Rates of all chemical processes are conventiona kPa/s un Therefore, you want to convert the 13 4 Expt 2. Decomposition of Hydrogen Peroxide-Worksheet View Review Mailings References Aa 1 AABBC 1.1 AABBCc 1.1.1 AatbCcl AabbCcDdfer AutbCcDdEe Ac -w Aa v Heading 2 Heading 3 Heading 1 No Spacing Normal A Data and Observations (Be mindful of units and significant figures) Initial rate Temperature Total Reactants Run (kPa/s) volume 0.0179 60.0 mL 21.8 °C 1 kPa/s mL 3.0% HO2+-5 mL 0.5 M KI+50- 5 mL H-O 60.0 mL 0.0196 21.8 C 2 kPa/s 10 mL 3.0 % H2O2 +-5_mL 0.5 M KI+45 mL H2O 60.0 mL 0.0514 21.8 C 3 5 mL 3.0 % H2O2+10 mL 0.5 M KI+-45 mL H2O kPa/s 0.0976 kPa/s 13.2 C 60.0 mL 4. 10 mL 3.0 % H2O 2 + -5 mL 0.5 M KI+ 45 mL HO Observations during the experiments: When mixing HO2, KI, and H2O together, the substances turned yellow color. After stirring, bubble started to form. The initial rate (in kPa/s) was calculated in the following way (brief description): Find pressure vs. time graph Select steepest 20 second segments Next analyze graph to fit curve then click linear, and finally pressure C Compared to run 1, run 2 uses twice as concentrated H2O), while keeping [KI] the same. How does the initial rate change? What does it telll us about order of the reaction by H2O1? The initial rate was 3-4 times more than rate two. It's first order reaction because rate depends on (United States) Focus Mailings Review View ences Ap AatibCcDdEe 1 AABBC AaßhCrDdFe 1.1 AABBCC 1.1.1 AaßbCel Normal No Spacing Heading 1 Heading 2 Heading 3 Analysis Initial rate [HO after mixing Rate constant Run after mixing (M/s) 2 3 4 ADVICE FOR FILLING IN THE TABLE. 1) Rates of all chemical processes are conventionally measured in M/s (molarity per second). Therefore, you want to convert the kPa/s units into M/s. Recall that the ideal gas law is: PV nRT C 2) Divide both parts of the ideal gas law by V R T s. What does molarity (moles over liters) per second equal to, based on your rearranged equation? 3) Use your finding above to calculate the first column of the analysis table. Use the literature value of 8.314 LkPa/Kmol for the ideal gas constant R. Show here how to calculate the run 1's value: The moment you mixed your 3 % ( by weight) H:O2 with other solutions, its concentration (underline) decreased/increased. You can always use M;Vi MaV to calculate molarity changes during dilutions. As long as you use same units for volume on both sides, formula will work, 3 % (by weight) H20, is the same as 0.88 M. Show here how to calculate the run 1's [HOa] value. 4) d States) Focus MacBook Pro 80 DC DII E5 F6 F7 FA

Answer 1

The '/s ' or per second term is present possibly to highlight the order of the reaction which will help you to solve for n in the PV=nRT equation.