using bond enthalpies - Problems 11 - 15

Go to Hess' Law - using bond enthalpies - Problems 1 - 10

Go to Hess' Law - using bond enthalpies

Go to Hess' Law - using two equations and their enthalpies

Go to Hess' Law - using three equations and their enthalpies

Go to Hess' Law - using four or more equations and their enthalpies

Go to Hess' Law - using standard enthalpies of formation

Back to the Termochemistry menu

**Problem #11:** Calculate the enthalpy change for the reaction of ethene and hydrogen, given the following bond energy values in kJ/mol:

H-H 436; C-H 412; C=C 612; C-C 348

**Solution:**

1) The chemical reaction is this:

C_{2}H_{4}+ H_{2}---> C_{2}H_{6}

List the bonds broken and the bonds made:

reactant bonds broken: four C-H bonds, one C=C bond, one H-H bond

product bonds made: one C-C bond, six C-H bonds

3) You can reduce that to this:

reactant bonds broken: one C=C bond, one H-H bond

product bonds made: one C-C bond, two C-H bonds

Note that four C-H bonds were removed from each side.

4) Hess' Law:

ΔH = Σ E_{reactant bonds broken}minus Σ E_{product bonds broken}ΔH = [(C=C) + (H-H)] - [(C-C) + (2) (C-H)]

ΔH = [612 + 436] - [348 + 824]

ΔH = -124 kJ

**Problem #12:** Calculate the C=C bond energy in ethene:

H_{2}C=CH_{2}(g) + H_{2}(g) --> H_{3}C-CH_{3}(g) ΔH = -138 kJ/mol

Bond enthalpies (kJ/mol): C-C = 348; H-H = 436; C-H = 412

**Solution:**

1) Hess' Law for bond enthalpies is:

ΔH = Σ E_{reactant bonds broken}minus Σ E_{product bonds broken}

2) Let's insert symbols, not numbers:

ΔH = Σ ([(C=C) + (4)(C-H) + (H-H)] minus Σ [(6)(C-H) + (C-C)]Do not forget that C-C bond! I initially forgot it when I solved this problem prior to formatting it for the web site.

3) Cancel 4 C-H bonds:

ΔH = Σ ([(C=C) + (H-H)] minus Σ [(2)(C-H) + (C-C)]

4) Put numbers in place and solve:

-138 = Σ ([(x) + (436)] minus Σ [(2)(412) + (348)]x = 598 kJ

**Problem #13:** The following two equations produce methane and ethane:

C + 4H ---> CH _{4}ΔH = -1652 kJ/mol 2C + 6H ---> C _{2}H_{6}ΔH = -2825 kJ/mol

(a) Calculate the bond enthalpy of a C-H bond.

(b) Calculate the bond enthalpy of a C-C bond.

**Solution:**

In the first equation, 4 C-H bonds are formed.-1652 kJ/mol divided by 4 = 413 kJ <--- that's the bond enthalpy of a C-H bond

Note that bond enthalpies are expressed as a positive value (energy put into the bond to break it), so I ignored the minus sign on the 1652 value.

For the second reaction, note that six C-H bonds are formed and one C-C bond is formed.

413 times 6 = 2478

2825 minus 2478 = 347 kJ <--- that's the bond enthalpy of a C-C bond

**Problem #14:** Calculate the mean bond enthalpy of the Si-F bond in SiF_{4}(g) given:

enthalpy of formation of SiF_{4}(g) = -1615 kJ mol¯^{1}

enthalpy of atomization of silicion = +456 kJ mol¯^{1}

enthalpy of atomization of fluorine = +79 kJ mol¯^{1}

Here is the Wikipedia entry for enthalpy of atomization.

**Solution:**

1) Let's write out what we are given using chemical equations:

Si(s) + 2F _{2}(g) ---> SiF_{4}(g)ΔH _{f}= -1615 kJSi(s) ---> Si(g) ΔH _{a}= +456 kJ(1/2)F _{2}(g) ---> F(g)ΔH _{a}= +79 kJ

2) Here's the reaction we want:

SiF_{4}(g) ---> Si(g) + 4F(g)The mean bond enthalpy is the energy required to break a bond, in this case one Si-F bond. Important point: the above reaction breaks four Si-F bonds. That will come into play below.

3) Adjust the given reactions as follows:

eq 1 - flip

eq 2 - leave unchanged

eq 3 - multiply by 4

4) Resulting in:

SiF _{4}(g) ---> Si(s) + 2F_{2}(g)ΔH = +1615kJ Si(s) ---> Si(g) ΔH = +456kJ 2F _{2}(g) ---> 4F(g)ΔH = +316 kJ

5) Add the three equations and their enthalpies to get:

+2387 kJ

6) That is the enthalpy to disrupt four Si-F bonds, so divide by 4 to get:

+596.75 kJwhich, to three significant figures, rounds off to +597 kJ

**Problem #15:** The decomposition reaction of tetrahedral P_{4} is as follows:

P_{4}(g) ---> 2P_{2}(g); ΔH = +217 kJ

If the bond energy of a single P-P bond is 200 kJ mol−1 what is the energy of the PP triple bond in P_{2}?

**Solution:**

Say you break all 6 P-P bonds in P_{4}, that 6 x 200 = +1200.1200 - 217 = 983 which is released when two P≡P bonds form.

So 983 / 2 = bond energy of a P≡P bond.

**Problem #16:** Given that a chlorine-oxygen bond in ClO_{2}(g) has an enthalpy of 243 kJ/mol, an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the standard enthalpy of formation of ClO_{2}(g) is 102.5 kJ/mol, use Hess's law to calculate the value for the enthalpy of formation per mole of ClO(g).

**Solution:**

1) This is our target equation:

(1/2)Cl_{2}(g) + (1/2)O_{2}(g) ---> ClO(g); ΔH = ???

2) Let's see what we know:

(a) we know an enthalpy:(1/2)Cl_{2}(g) + O_{2}(g) ---> ClO_{2}(g); ΔH = 102.5 kJ(b) and we know two bond enthalpies:

Cl-O = 243 kJ/mol

O=O = 498 kJ/mol

3) The first step is to use a bond enthalpy calculation to determine the bond enthalpy of the Cl-Cl bond in Cl_{2}

ΔH = Σ E_{reactant bonds broken}minus Σ E_{product bonds broken}102.5 = [(1/2)(x) + 498] - [(2)(243)]

102.5 = [(1/2)(x) + 12

x = 181 kJ

4) Now, use a second bond enthaphy calculation for the equation in step 1 above:

ΔH = Σ E_{reactant bonds broken}minus Σ E_{product bonds broken}x = [(1/2)(181) + (1/2) (498)] - 243

x = 96.5 kJ

A different approach to solving this problem may be found here.

Two links to Yahoo Answers bond enthalpy problems:

Calculate the bond enthalpy of the Cl—Cl bond.

Calculate the bond enthalpy for the N-H bond in NH_{3}

Go to Hess' Law - using bond enthalpies - Problems 1 - 10

Go to Hess' Law - using bond enthalpies

Go to Hess' Law - using two equations and their enthalpies

Go to Hess' Law - using three equations and their enthalpies

Go to Hess' Law - using four or more equations and their enthalpies