For the reaction, Delta H^0 = 285.4 kJ and Delta S^0 = 137.4 J K. What is Delta

Question

For the reaction, Delta H^0 = 285.4 kJ and Delta S^0 = 137.4 J K. What is Delta G^0 in kJ an 95 degree C? 3 O_2 (g) rightarrow 2 O_5 (g) a) -285.4 b) 336.0 c)272.3 d) 298.5 e) 234.8 Given the standard molar entropies listed beneath each compound, calculate Delta for this reaction in 25 degree C C_2 H_4 (g) + 3 O_2 (g) rightarrow 2 CO_2 (g) + 2H_2 O(g) S in mol-K 219.4 205.0 213.6 60.91 a) -140.9 J b) +140.9 J c) + 332.6 J d) -267.4 J e) + 207.8 What are the coefficients of the permanganate ion and iodide ion (the reactions, in order) when the following equation is balanced? MnO_4^- + I^- rightarrow Mn^2+ + I_2 a) 1 and 1 b) 4 and 2 c) 1 and 4 d) 2 and 3 e) 2 and 10 Which of the following is correct? a) A catalyst is used up when it takes part in a reaction. b) a catalyst changes the mechanism of a reaction c) A catalyst shifts an equilibrium to give more product. d) A catalyst raises the activation energy of a reaction e) A catalyst raises the forward rate constant but decreases the reverse rate constant.

Explanation / Answer

35)
Ho = 285.4 KJ
So = -137.4 J/K = -0.1374 KJ/K
T = 95 oC = (95 + 273) K =368 K

use:
Go = Ho - T*So
= 285.4 - 368*(-0.1374)
= 336.0 KJ

Answer: b

I am allowed to answer only 1 question at a time

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.