Calculate Delta G degree for the combustion of propane. -2073.1 kJ -1387.3 kJ -5

Question

Calculate Delta G degree for the combustion of propane. -2073.1 kJ -1387.3 kJ -598.5 kJ 2073.1 kJ Calculate the equilibrium constant at 25 degree C for the reaction of methane with water to form carbon dioxide and hydrogen. The data refer to 25 degree C. 8.2 Times 1019 0.96 0.58 1.2 Times 10-20 1.4 Times 10-46 Consider the following balanced redox reaction Which of the following statements is true? Mn2+ (aq) is the oxidizing agent and is reduced. Mn2+ (aq) is the oxidizing agent and is oxidized. Mn2+ (aq) is the reducing agent and is oxidized. Mn2+ is the reducing agent and is reduced. Manganese does not change its oxidation number in this reaction. Consider the following redox equation When the equation is balanced with smallest whole number coefficients, what is the coefficient for OH- (aq) and on which side of the equation is OH-(aq) present? 4, reactant side 4, product side 6, reactant side 6, product side none of the above

Explanation / Answer

25) products - reactants = (-2097.6) - (-24.5) = -2073.1 kJ. So A is your answer

26) deltaG = (-394.4) + 4(0) - (-50.91) - 2(-228.6) = 113.6 kJ
deltaG= -RT ln K
Keq= e-113.6 / ( .008314*298) = 1.2 * 10-20So your answer is D

27) Mn is 2+ on left side, but 4+ on right side. This means it is oxidized. S goes from +7 on left side to +3 on right side. This means it is reduced. Since it is gaining the electrons from Mn2+, it becomes the oxidizing agent. This makes Mn2+ the reducing agent. C is your answer

28)
Mn(OH)2 + 4 OH- = MnO42- (aq) + 2 H2O + 2 e-
MnO4- (aq) + e- = MnO42- (aq)
======================================...
Mn(OH)2 + 2 MnO4- (aq) + 4 OH- --> 3 MnO4- (aq) + 2 H2O

So 4 on the reactant side. your answer is A

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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