Using the Page 8 of 9 for the following bond enthalpies provided below, determin

Question

Using the Page 8 of 9 for the following bond enthalpies provided below, determine the value of delta H_nm reaction? 2 C_2H_2 (g) + 5 O_2 rightarrow 4 CO_2 (g) + 2 H_2O(e) H - C = C - H O = O O = C = O H - O - H Bond energies (KJ mol^-1): C - H: 414; H - O: 464; C = C: 612 O = O: 498; C = O: 715; C = C: 820 705 kJ -1070 kJ -1790 kJ -2206 k J -2618 kJ Calculate the bond energy of the F-F bond, given the following data: Bond energies (kJ mo^-1): C - F: 485 delta H degree_f (KJ mol^-1): CF_3(g): -508.5 C(atom, g): 717 76 kJ mol^-1 153 kJ mol^-1 230 kJ mol^-1 831 k mol^-1 1445 mol^-1 Determine the delta H degree _f of acetaldehyde, CH_3CHO(g) in kJ mol^-1, using the following data. Average bond energies (kJ mol^-1): (C - H) 414 (C - C) 347 (C = O) 715 Standard enthalpies of formation (delta H degree _f, kJ mol^-1): C(atomic, g): +717 H(atomic, g): +218 O(atomic, g): +249

Explanation / Answer

Q25

C-F = 485

CF3 = -508.5

C(s) = 717

so..

C(s) + F2(g) = CF3(g)

balance:

2C(s) + 3F2(g) = 2CF3(g)

now...

Hreaciton = Hproducts - Hreactants

C-F bonds = 3

F-F bonds = 1

C(s) = 0 bonds...

so

2C(s) + 3F2(g) = 2CF3(g)

HRxn = 2*(3*485) - 3*(F-F)

HRxn = -508.5

-508.5= 2*(3*485) - 3*(F-F)

F-F = ((-508.5) -6*485 ) /3 = 1139.5

So nearest answer is E

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