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Q.Calculate the work done (in J) when 4.5 g of H 2 O 2 reacts against a pressure of 1.0 atm at 25°C 2H 2 O 2 ( l ) —→ O 2 (g) + 2 H 2 O ( l ) (a) -1.63 x 10 2 (b) 4.5 x 10 2 (c) 3.2 x 10 2 (d) - 6.1 x 10 2

Q.Calculate the work done (in J) when 4.5 g of H2O2 reacts against a pressure of 1.0 atm at 25°C 2H2O2 (l) —→ O2 (g) + 2 H2O (l) 
 
(a) -1.63 x 102 (b) 4.5 x 102 (c) 3.2 x 102(d) - 6.1 x 102

Grade:11

1 Answers

Arun
25750 Points
5 years ago
Molecular mass of H2O2=34 gm/mol
4.5 gm of H2O2=4.5/34=0.132 mol
So, from the reaction, 0.132 moles of H2O2 reacts and produces 0.132/2=0.066 moles of O2.
Taking volume of liquid negligible,
Change in volume during process assuming ideal gas behaviour=(0.132-0.066)*22.4L=1.4784 L
So, total work done=PdV=1.013*10^5*1.4784*10^-3=1.4976*102�J.
Which is done on the system, so negative.=-1.4976*102�J

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