Answer to Question #142187 in Physical Chemistry for Hatsh

Question

A 1litre vessel which is equipped with a movable piston is filled with a 2M aqueous solution of h2O2.the h2O2 decomposes to h2O and O2 is a first order process with half life 5 hours at 300K.as gas formed the piston moves upward against external pressure of 1atm.calculate work done by the gas from the start of 6hours till the end of 15 hours

Answer 1

2H₂O₂ = 2H₂O + O₂

k = ln(2)/5 = 0.139 h^-1

ln(C₀/C) = kt

6 hours:

ln(2/C) = 0.139*6

C = 0.869

15 hours:

ln(2/C) = 0.139*15

C = 0.249

$\Delta$C_H2O2 = 0.869-0.249 = 0.620 M

$\Delta$n_H2O2 = 0.620 mol

$\Delta$n_O2 = 0.310 mol

$\Delta$V_O2 = $\Delta$n_O2*R*T/p = 0.310*8.314*300/101325 = 0.00763 m³

W = p*$\Delta$V_O2 = 101325*0.00763 = 773 J