Question

Question 47: Use the bond energies value in table 13.6 to estimatefor each of the following reactions in the gas phase

1. ${\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{CI}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{2}{\mathbf{HCL}}_{\left(g\right)}$
   
2. $\mathbf{N}\mathbf{\equiv }{\mathbf{N}}_{\left(g\right)}\mathbf{+}\mathbf{3}{\mathbf{H}}_{\left(2g\right)}\mathbf{\to }\mathbf{2}{\mathbf{NH}}_{\mathbf{3}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$
   
3. $\mathbf{H}\mathbf{-}\mathbf{C}\mathbf{\equiv }{\mathbf{N}}_{\left(g\right)}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }$
   

4.

$\mathbf{+}\mathbf{2}{\mathbf{F}}_{\mathbf{2}\left(g\right)}\mathbf{\to }\mathbf{N}\mathbf{\equiv }\mathbf{N}\mathbf{+}\mathbf{4}{\mathbf{HF}}_{\left(g\right)}$

Short answer

Bond energy is defined as the energy required to break a bond.

$\to$enthalpy change during a reaction takes place as

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{product}}-\sum ∆{\mathrm{H}}_{\mathrm{reactant}}$

as a bond breaks and a product, bond is formed during a reaction.

Step-by-step solution

Sub-part a:∆H for H2g+CI2g→2HCI(g)

Step 1: enthalpy change for a reaction

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{product}}-\sum ∆{\mathrm{H}}_{\mathrm{reactant}}$

Step 2: Bond break in Reactant side with Bond energy

1mol H-H $\to$ 436KJ/mol

1mol Cl-Cl $\to$242KJ/mol

Step 3: Bond formed in product side with Bond energy

2mol HCl $\to 2\times 431$KJ/mol = 862KJ/mol

Step 4:

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{p}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}$

= -862KJ +678KJ

= -184KJ

Hence$∆\mathrm{H}$ Reaction for HCl formation is -184KJ

Sub-part b: ∆HforN≡Ng=+3H2g→2NH3g

Step 1: enthalpy change for a reaction

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{product}}-\sum ∆{\mathrm{H}}_{\mathrm{reactant}}$

Step 2: Bond break in the Reactant side

1mol $\mathrm{N}\equiv \mathrm{N}$$\to$945KJ/mol

3mol H-H $\to 436\times 3$=1308 KJ/mol

2253KJ

Step 3: Bond formed in product side with Bond energy

6mol of N-H bond $\to 6\times 389$KJ = 2334KJ

Step 4:

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{p}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}$

= 2334 - 2253

= -81 KJ

Sub-part c:∆H  for H-C≡N(g)+2H(2g)→

Step 1: enthalpy change for a reaction

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{product}}-\sum ∆{\mathrm{H}}_{\mathrm{reactant}}$

Step 2: Bond break in the Reactant side

1mol H-C$\to 1\times 413\sum _{}^{}∆{\mathrm{H}}_{\mathrm{R}}$=413+891+872

1mol$\mathrm{C}\equiv \mathrm{N}\to 1\times 891$= 2176KJ

1mol H-H$\to 2\times 436$

Step 3: Bond formed in product side with Bond energy

3mol C-H $\to 3\times 413$= 1239

2mol N-H $\to 2\times 391$ = 782

1mol C-N $\to 1\times 305$=305

$\sum _{}^{}∆{\mathrm{H}}_{\mathrm{R}}=$2326

Step 4:

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{p}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}$

= 2326 - 2176

= -150 KJ

Sub-part d:  ∆H for+2Fe2g→N≡N+4HF(g)

Step 1: enthalpy change for a reaction

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{product}}-\sum ∆{\mathrm{H}}_{\mathrm{reactant}}$

Step 2: Bond break in Reactant side

4mol N-H $\to 4\times 391$= 1564

1mol N-N$\to 1\times 160$= 160

1mol C-N $\to 2\times 154$=308

2032KJ

Step 3: Bond formed in product side with Bond energy

1mol $\mathrm{N}\equiv \mathrm{N}\to 1\times 941$= 941

2mol H-F= 2260

3201KJ

Step 4:

$∆{\mathrm{H}}_{\mathrm{reaction}}=\sum _{}^{}∆{\mathrm{H}}_{\mathrm{p}}-\sum ∆{\mathrm{H}}_{\mathrm{R}}$

= 3201-2032

= -1169 KJ

Hence enthalpy of the reaction is -1169KJ