How much heat (in joules) is liberated when 1.170L of ozone at 18 degrees celsius and 62.8 mm Hg decomposes to ordinary oxygen, O2?
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wow thats quite a question they gave you there but is here how to solve it:
W(sys)=-P(delta)V is the equation you use once you get the volume change and in order to get the volume change we need to use the ideal gas law:
PV=nRT, so convert 62.8mmHg to atm (atmospheres)
V=(1mol)(.08206)(291K)/(.0826atm)=291L… now input into P(delta V)
1.17L - 291L= -289.83L now divide into 1000
-(101325Pa x (-291L/1000m^3)=29367J of work done on surroundings so -29367 J of work done to decompose to O2
W(sys)=-P(delta)V is the equation you use once you get the volume change and in order to get the volume change we need to use the ideal gas law:
PV=nRT, so convert 62.8mmHg to atm (atmospheres)
V=(1mol)(.08206)(291K)/(.0826atm)=291L… now input into P(delta V)
1.17L - 291L= -289.83L now divide into 1000
-(101325Pa x (-291L/1000m^3)=29367J of work done on surroundings so -29367 J of work done to decompose to O2