4C.3

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Ellis Song 4I
Posts: 102
Joined: Thu Jul 11, 2019 12:17 am

4C.3

Postby Ellis Song 4I » Thu Jan 30, 2020 6:09 pm

Calculate the final temperature and the change in enthalpy when 765 J of energy is transferred as heat to 0.820 mol Kr(g) at 298 K and 1.00 atm (a) at constant pressure; (b) at constant volume. Treat the gas as ideal.

For part a the solutions manual says q=(5.025 g/83.80 g/mol)(25C - 97.6C)(20.8 J/molC) = -90.6 J.

Where did the 5.025g come from? And how did they find that temperature change?

Ipsita Srinivas 1K
Posts: 50
Joined: Mon Jun 17, 2019 7:24 am

Re: 4C.3

Postby Ipsita Srinivas 1K » Thu Jan 30, 2020 10:26 pm

Okay, the temperature bit super weird because the actual textbook (not the solutions manual) has all the odd-numbered answers and it's only given a temperature change for part b = 373 K.
I think the 5.025 g is the moles of Kr given converted to grams, over the molar mass of Kr, which is approximately 83 g/mol. I think the manual is just trying to show a molar conversion because it's using the specific heat capacity in grams, not the molar heat capacity.


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