Question

(2). Consider benzene in the gas phase, CaH&(g). Use the heat of formation, AH, values below to answer the questions. (20 points) дня (/mо) Substance H(g) C(g) С.Н.A(9) 217.94 718.4 82.9 (a) What is the standard enthalpy change for the reaction that represents breaking all the bonds in gaseous benzene, C.Ha(g)? (6 points) (b) What is the chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds in gaseous benzene, CH6(? Indicate the phase of each species in your answer. (4 points) (e) The average bond enthalpy for C-H is 413 kJ/mol. 413 kJ of energy is required to break a mole of CH into atoms: CH(g)C(g)+H(g). AH-413 kJ Using this information, and your answer from (a), calculate the enthalpy change of the reaction from (b). That is, calculate the energy required to break only the carbon-carbon bonds in benzene. (10 points)

Answer 1

The standard enthalpy change for this reaction is expressed by the following formula:

ΔH^o_reaction= ∑m⋅ΔH_f^o (products) − ∑n⋅ΔH_f^o(reactants) ⋅⋅⋅⋅⋅⋅⋅(1)

Where, ΔH_f^o (products) = standard enthalpy of formation of products, ΔH_f^o(reactants) =  standard enthalpy of formation of reactants, m is the total moles of products and n is the total moles of reactants.

Breaking bond rxn can be written as:

C₆H₆ (g) → 6C (g) + 6H (g)

ΔH^o_reaction=  [6 X ΔH_f^o (H(g)) + 6 X ΔH_f^o (C(g))] − [ΔH_f^o (C₆H₆ (g))]

= [6 moles X (217.94 kJ/mol) + 6 moles X (718.4kJ/mol)] − [1 mole x (82.9kJ)]

= (1307.64 + 4310.4 - 82.9) kJ

ΔH^o_reaction= 5535.14 kJ

(b) The equation for reaction is: C₆H₆ (g) $\rightarrow$ 6 CH (g) (2)

(c) CH (g)    $\rightarrow$ C (g) + H (g)

For this rxn, ΔH^o_reaction= 413 kJ

Using (1)

ΔH^o_reaction=  [ ΔH_f^o (H(g)) + ΔH_f^o (C(g))] − [ΔH_f^o (CH(g))]

413 = 217.94 + 718.4 - [ΔH_f^o (CH(g))]  

[ΔH_f^o (CH(g))] = 523.34 kJ

So, calculating enthalpy change of reaction in (2):

ΔH^o_reaction=  [6 X ΔH_f^o (CH(g)) ] − [ΔH_f^o (C₆H₆ (g))]

= [6 X 523.34] - 82.9 = 3140.04 - 82.9

ΔH^o_reaction = 3057.14 kJ

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