Question

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

Answer 1

Answer:

option d= -440 kj

Explanation:

chemical equation:

CH4 + 4Cl2 → CCl4 + 4HCl

Given data:

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