d luck. Question 1 5 pts Hydrogen peroxide (FW 34.0 g/mol) decomposes irreversib

Question

d luck. Question 1 5 pts Hydrogen peroxide (FW 34.0 g/mol) decomposes irreversibly to gaseous water and oxygen as shown by the equation below: 2H202(s) 2H2O() +02(g) If 4.000g of hydrogen peroxide was placed inside a balloon, which expanded under standard state conditions (25 °C, 1.0 atm) how much work (in joules) would be performed by the reaction (aka the system)? Assume the volume change of solid hydrogen peroxide can be disregarded and that the reaction proceeds to completion. DQuestion 2 5 pts MacBobk

Explanation / Answer

First we have to calculate the moles of O2 gas produced and then, volume of N2 gas.

molar mass of H2O2 = 34.01 g/mol

molar mass of O2 = 32 g/mol

2H2O2(s ) 2 H2O(l) + O2(g)

2 mol 1 mol

2 x 34.01 g = 68.02 g 1 mol

4.0 g ?

Then,

? = (4.0 g/ 68.02 g ) x 1 mol of O2

= 0.0588 mol of O2

Therefore,

moles of O2 , n = 0.0588 mol

By using ideal gas equation PV = nRT, we have to calculate the volume of O2 at 1.00 atm and 25 °C.

P = 1 atm

T = 25oC = 25 + 273 K = 298 K

R = Ideal ags constant = 0.0821 L.atm/K/mol

Then, PV = nRT

V = nRT/P

= 0.0588 mol x 0.0821 L.atm/K/mol x 298 K/ 1 atm

= 1.44 L

Hence, volume of O2 = 1.44 L

Then,

work = - PV

= - 1 atm x 1.44 L

= - 1.44 L.atm

= -1.44 x 101.3 J [ 1 L.atm = 101.3 J]

= - 145.87 J

Therefore,

work performed by the reaction = - 145.87 J

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