16) Given the following reactions Fe2O3 (s) + 3CO (s) 2Fe(s) + 3CO2 (g) AH--28.O

Question

16) Given the following reactions Fe2O3 (s) + 3CO (s) 2Fe(s) + 3CO2 (g) AH--28.Okj 3Fe (s) + 4CO2 (s) 4CO (g) + Fe304 (s) AH-+12.5 kJ the enthalpy of the reaction of Fe203 with CO 3 Fe2O3 (s)-CO (g) CO2 (g) + 2Fe3Og (s) A) -155 B) 40.5 C)-109 D) -59.0 17) Use the following thermochemical equations C2H2 (g) + 5/202 (g) 2CO2 (g) + H2O (g) AH--1300 kj C2H6 (g) + 7/202 (g) -+2CO2 (g) + 3H20(g) AH=-1560 kJ H2 (g) + 1/202 (g) H2O (1) AK -286kJ to calculate Ho for the following reaction: C2H2 (g) + 2H2 (g) C2H6 (g) A) -2860 kJ B) +1560 k C)-312 kJ D) -3146 k) 18) Given the following equations: 2 H2 (g) + O2 (g) 2H2O (l) N2 (g) + O2 (g) 2 NO (g) N2 (g) + 3 H2 (g) 2 NH3 (g) AH =-571.6 kJ AH = + 180.5 kJ AH =-92.22 kJ Determine the enthalpy change (AH) for the following reaction 2 NO (g) + 5 H2 (g) 2 NH3 (g) + 2 H2O (1) A) -844.3 kJ/mol B) 844.3 kJ/mol O)-483.3 kJ/mol D) 483.3 kJ/mol

Explanation / Answer

The solutions are as follows

16) We have to calculate for,

3Fe2O3(s) + CO(g) ------> CO2(g) + Fe3O4(s)

To get this equation, we multiply the first reaction equation with 3 and the second with 2. Adding both the reaction equation will give is the net required equation.

Thus,

enthalpy of the reaction = (3 x -28.0) + (2 x 12.5) = -109 kJ

Thus the correct answer is C) -109

17) In this reaction we have to multiply the third given equation with 2, reverse the second equation and add all three, will give us the net reaction equation as,

C2H2(g) + 2H2(g) -------> C2H6(g)

Thus, delta Horxn = -1300 + (2 x -286) + 1560 = -312 kJ

Thus the correct answer is C) -312 kJ

18) To get the net reaction we have to reverse the second equation and then add all three euqation will give us the net reaction as,

2NO(g) + 5H2(g) -------> 2NH3(g) + 2H2O(l)

thus, enthalpy chage for the reaction = -180.5 + (-571.6) + (-92.22) = -844.3 kJ/mol

thus the correct answer is A) -844.3 kJ/mol

19) Delta Horxn = delta Hoproducts - delta Horeactants

                          = -2346 + (2 x -285.9) - (-986.6 + 2 x -900.4)

                          = -130.4 kJ/mol

Thus the correct answer will be D) -130.4

20) As seen in 19),

Delta Horxn = delta Hoproducts - delta Horeactants

                    = (4 x 90 + 6 x -286) - (4 x -46 + 5 x 0)

                    = -1172 kJ/mol

thus the correct answer will be B) -1172

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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