## thermal chem

I need some help with thermal chem one problem in particular.
When 1.50 g of Ba(s) is added to 100.00 g of water in a container open to the atmosphere, the reaction shown below occurs and the temperature of the resulting solution rises from 22.00C to 33.10C. If the specific heat of the solution is 418 J/(g*C) calculate delta H for the reaction, as written.
Ba(s)+2H2O(l)->Ba(OH)2(aq)+H2(g)
some explanation would be greatly appritiated.

Ps sorry about the spelling :oops: i'm pretty good at chem but the language thing escapes me. :?

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Well, the idea here is to be able to quantify the change in energy of the reaction. Well, the problem implicitly states that all the energy evolved from the reaction gets transfered in the form of thermal energy (hence, the rise in temperature of the solution from 22.00C to 33.10C). Well, the specific heat of the solution is 418 jouls per gram of solution per degrees C. To find the change in energy, we multiply the specific heat by the mass of the solution and by the change in temperature of the new solution.

Remember that by adding 1.50 grams of Barium, we have increased the mass of the solution, which you should be able to do from stoichiometry. The mass of the new solution is 101.47 grams.

Therefore, the change in energy due to the reaction is -(418 J/g*C)(33.1C-22.0C)(101.47 g) = -470.8 kJ. However, the question asks for delta H for the reaction, as written. Therefore, we divide by the number of moles of Barium chucked in water to arrive at the answer (molar enthalpy). From your calculations, you should get 0.0109 moles.

We divide through, and arrive at delta H = -470.8 kJ/0.0109 moles = 4,310 kJ/mol

### dumb it down pleeeeze

hey thank you for the help but not to be dificult or anything but i'm not all that good on thermal chem (if you haven't noticed) could you dumb it down a little more for me pleeeze. o and i furgot the specific wieght was supposed to be 4.18 not 418 srry about the typo. :oops:

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