Question

please help! I only have barometer and total value, I don't know how to do the rest

424 Report Sheet.Rates of Chemical Reactions It: Rate and Order, B. Effect of Temperature BATH TEMPERATURE Burer read Vol. of o: ing imL) imL) Time ml. Total volume of O, collected Barometric pressure mm Hg Vapor pressure of water at bath temperature (see Appendix L) Slope mm Hg times faster Compared with the rate found for solution 1, the rate is Using the ideal gas law, calculate the moles of O, collected (show calculations) mol O Based on the moles of O, evolved, calculate the molar concentration of the original 3%. H,0, solution (show calculations)

barometer is 752 mmHg

Answer 1

Volume of O2 collected = 16.4 ml

Barometric Pressure = 752 mmHg

Vapou pressure of water at 33.1 0C = 37.7 mm Hg

Temperature of bath = 33.1 oC= 306.25 K

P=752 mmHg

$H_{2}O_{2} \rightarrow H_{2}O+\frac{1}{2}O_{2}$

rate of reaction = $= \frac{\Delta [O_{2}]}{total time \times \frac{1}{2}}$

time	volume of o2 collected
0	0
29	2
50	4
70	6
87	8
105	10
124	12
142	14
260	16.4

volume of o2 collected vs time 25 y = 0.068x+1.430 20 15 volume ml 10 Series1 Linear (Series1) 5 0 0 100 200 300 time second

rate of reaction =0.126 ml/s

As data for solution 1 is not given we cant compare it.

Ideal gas equation

PV=nRT

P=752-37.7=714 mm Hg=0.94 atm

volume = 16.4 ml = 0.0164 l

R=0.0821 l atm/mol/K

n=?

T= 306.25 K

keeping value in the above equation and calculating n

n=0.0006 moles of O2

1/2 moles of O2 is produced from = 1mole of H2O2

1 mole of O2 is produced from = 2 moles of H2O2

0.0006 moles of O2 produced from= 0.0012 moles of H2O2

As volume of initial H2O2 is not given let us consider it 10 ml

mass of h2O2 =0.012 X 34 =0.48 g

3% H2O2 means

3 g h2o2 is present in= 100 ml solution

0.48 g is present in = $=\frac{100\times 0.48}{3}= 16 ml$

$molarity = \frac{n\times 1000}{volume}=\frac{0.012\times 1000}{16}=0.75 M$