Textbook question 4J.17






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Paige Lee 1A
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Joined: Sat Sep 07, 2019 12:16 am

Textbook question 4J.17

Postby Paige Lee 1A » Wed Feb 12, 2020 12:42 pm

Could someone please explain why for B and C you know that the reactions are spontaneous everywhere and nonspontaneous everywhere respectively? Why don't you calculate the maximum temperature cut off? Why are parts B and C different from part A of this question?

Assume that DH8 and DS8 are independent of temperature and use data in Appendix 2A to calculate DG8 for each of the
following reactions at 80. 8C. Over what temperature range will each reaction be spontaneous under standard conditions?
(a) B2O3(s) 1 6 HF(g) ¡ 2 BF3(g) 1 3 H2O(l)
(b) CaC2(s) 1 2 HCl(aq) ¡ CaCl2(aq) 1 C2H2(g)
(c) C(s, graphite) ¡ C(s, diamond)

Connie Chen 1E
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Joined: Mon Jun 17, 2019 7:24 am

Re: Textbook question 4J.17

Postby Connie Chen 1E » Thu Feb 13, 2020 7:50 pm

You know that reaction B is spontaneous everywhere because Δ H is calculated to be negative and Δ S is calculated to be positive. By using the Gibbs free energy formula, Δ G = Δ H − T Δ S, we will get a negative number minus a positive number for any temperature, which will only result in negative answers, and a reaction is spontaneous when Δ G is negative.

Similarly for reaction C, Δ H is positive and Δ S is negative, so we get a positive number minus a negative number for any temperature, which results in positive Δ G's. This means the reaction will not be spontaneous.

JasonLiu_2J
Posts: 109
Joined: Sat Aug 24, 2019 12:17 am

Re: Textbook question 4J.17

Postby JasonLiu_2J » Fri Feb 14, 2020 12:15 pm

Whether or not you can calculate the temperature cutoff depends on whether ∆H and ∆S are positive or negative. For the reaction, when ∆H is positive and ∆S is negative, then you know that ∆G must be a positive value, thus making the reaction nonspontaneous no matter the temperature (temperature has to be a positive value since you can't have T<0K, so only the signs of ∆H and ∆S are considered). Similarly, when ∆H is negative and ∆S is positive, you see that ∆G has to be a negative no matter the temperature. You can only calculate the temperature cutoff for spontaneous vs nonspontaneous when ∆H and ∆S have the same signs, as the value of the temperature will influence whether ∆G becomes negative or positive. If both are positive, you should expect a high temperature since you want T∆S to be greater than ∆H. If both are negative, you would expect a lower temperature, since you want ∆H to be a greater absolute value than that of T∆S. Hope this helps!


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