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25 January, 07:52

Using the bond energies provided below, calculate δh° for the reaction ch4 (g) + 4cl2 (g) → ccl4 (g) + 4hcl (g) bond energies: c-h = 413 kj/mol, cl-cl = 243 kj/mol, c-cl = 339 kj/mol, h-cl = 427 kj/mol

a. 1422 kj

b. 110 kj

c. 440 kj

d. - 440 kj

e. - 110 kj

+1
Answers (1)
  1. 25 January, 07:55
    0
    option d = - 440 kj

    Explanation:

    chemical equation:

    CH4 + 4Cl2 → CCl4 + 4HCl

    Given dа ta:

    Bond broken energies:

    C-H = 413 Kj/mol

    Cl-Cl = 243 Kj/mol

    Bond formation energies:

    C-Cl = 339 Kj/mol

    H-Cl = 427 Kj/mol

    Formula:

    δh° = ∑n (bonds broken) - ∑m (bonds formation)

    Solution:

    Total bonds broken energy:

    C-H = 413 Kj/mol 4*413 Kj/mol = 1652 Kj/mol

    Cl-Cl = 243 Kj/mol 4*243 Kj/mol = 272Kj/mol

    total = 2624 Kj/mol

    Total bond formation energy:

    C-Cl = 339 Kj/mol 4*339 Kj/mol = 1356 Kj/mol

    H-Cl = 427 Kj/mol 4*427 Kj/mol = 1708 Kj/mol

    total = 3064 Kj/mol

    Now we will put the values in formula to calculate the δh° for reaction.

    δh° = ∑n (bonds broken) - ∑m (bonds formation)

    δh° = (2624) - (3064)

    δh° = - 440 kj
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