O. Givan the following data, calculste the nomal boiling point for fomic acid HC

Question

O. Givan the following data, calculste the nomal boiling point for fomic acid HCOOH). 410 363 130.0 250.9 HCOOHg) a) 0.39 *C b) 389*0 c) 662 C d) 279 -C e 116 °C 1 1 . For which of the following processes would ASbe expected to be most positive? 12. Which of the following statamants is (are) always true? L In order for a process to be spontaneous, the entropyofthe universe must ncrease IL A system cannothave both energy disorder and positional disorder LIL IV. Sis zero for elements in their standard states a) I unix = 13. For the reaction CLOg)-yo,( )2coAs), AF-126.4 kDmol and AS--74.9 JK mol At 373°C, what is 4 a) 154.3 kJmol b) 48.5Imol c 174.81 mol d) 78.0kImol e 156.6kImol

Explanation / Answer

10. Answer is e. 116oC

Boiling: HCOOH(l) --> HCOOH(g)

H of boiling:
H of product - reactant = -363 kJ - ( -410 kJ) = 47kJ

S of boiling:
S of product - reactant = 250.9 J/K - (130 J/K) = 120.9 J/K

the formula:
G = H - TS

when at equilibrium ( melting or boiling points), becomes:
zero = H - TS

this rearranges to:
T = H / S

T = 47,000 joules / 120.9 joules

T = 388.75 Kelvin

388.75 Kelvin - 273 = 115.75o Celsius = 116oC = T = Normal Boiling Point of Formic Acid

11. Answer is d. 2NH4NO2(s) -------> 2N2(g) + O2(g) + 4H2O(g)

12. Answer is a. I

13. Answer is d. 78.0 kJ/mol

G = H - TS = 126.4kJ/mol - (646K)(0.0749kJ/Kmol) = 78.0kJ/mol

14. Answer is d. 2.72 * 1042

G = Gf(Product - reactant) = -32.88-209.2 = -242.08 kJ/mol

Keq = e-G/RT = e-((-242.08)/0.008314 * 298) = 2.72 * 1042

15.Answer is d. -9.90kJ

G = -RTlnKeq = -0.008314 * 700 * ln (5.48) = -9.90 kJ

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

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