8.49
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8.49
How would you go about doing number 49 for the chapter 8 homework? "Oxygen difluoride is a colorless, very poisonous gas that reacts rapidly with water vapor to produce O2, HF, and heat: OF2(g)+H2O(g)--> O2(g)+2HF(g) deltaH=-318J What is the change in internal energy for the reaction of 1.00 mol OF2?"
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Re: 8.49
You use the equation deltaU=q+w. You can use the given deltaH for q, and use the equation w=-PdeltaV=-deltanRT to find w. This gives you deltaU=q+w=deltaH-PdeltaV=-318kJ-2.48kJ=-320kJ.
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Re: 8.49
For this problem, you must look at the balanced equation and determine the change in moles, delta(n)=nfinal-ninitial. In this case, nfinal would be 3 moles since there are 3 moles on the products side of the equation, and ninitial would be 2 moles, as there are two moles on the reactants side of the equation. Therefore, delta(n)=3-2=+1 mol. Then you use the relationship between delta(H) and delta(U): delta(H)=delta(U)+delta(n)gas*R*T. You know that delta(H)=-318 kJ as listed in the problem for the reaction. Now you isolate that relationship for delta(U) to get -320 kJ as your change in internal energy!
Let me know if you need any further elucidation regarding any aspect of this problem!
Sunny
Let me know if you need any further elucidation regarding any aspect of this problem!
Sunny
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