Homework 8.93b

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Homework 8.93b

Postby OliviaShearin2E » Wed Jan 17, 2018 2:51 pm

(a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25 C and 1.00 bar.
(b) Using data in Appendix 2A, calculate the standard enthalpy of the reaction. (c) Calculate the change in internal energy, U , of the system.

The equation I wrote for this problem is 2C2H6(g)+ 15O2(g) → 12CO2(g) + 6H2O(g).
In part (a) of 8.93, I got the correct answer by assuming that H2O was produced in gas form, but when I tried part (b) using the enthalpies of formation method I got the wrong answer. I replaced the enthalpy of formation for H2O(g) with the enthalpy of formation for H2O(l) and then got the correct answer. Why does part (b) assume the H2O is liquid rather than gas? Thank you!

Curtis Tam 1J
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Re: Homework 8.93b

Postby Curtis Tam 1J » Wed Jan 17, 2018 7:01 pm

I think it's because at 25 degrees Celsius, H2O would be in the liquid state.

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