Answer: 5535.1KJ/mol
Explanation:The equation for breaking of all the bonds in the benzene is written as follows:
C_6 H_6(g) --> 6C(g) +6H (g)
∆H_rxn= ∑moles of product X∆H_PRODUCT -∑moles of reactants x ∆H_REACTANT
={6mols X ∆H_°f(C) + 6mols X ∆H_°f(H)} –{1 mol ∆H_°fC_6 H_6 (g)
={6 X 217.74+ 6X 718.4} – {82.9}KJ/mol
= 5618.04 – 82.9 KK/mol
=5535.1KJ/mol
b) The equation for the breaking of the carbon-carbon bonds in benzene is illustrated below as
C_6 H_6(g) --> 6C---H (g)
A) The standard enthalpy change for the reaction is ; 5535.1 KJ/mol
B) The chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds is : C₆H₆(g) --> 6C---H(g)
A) The standard enthalpy change for the reaction which represents breaking of all bonds
Reaction equation ; C₆H₆(g) --> 6C(g) + 6H (g)
∆H = ∑ ( moles of product * ∆Hpro ) - ( moles of reactant * ∆H reactant )
= ( 6 * 217.94 + 6 * 718.4 ) – ( 82.9) KJ/mol
= ( 5618.04 ) - ( 82.9 ) = 5535.1 KJ/mol
B) The chemical equation for the reaction that corresponds to breaking the carbon-carbon bonds
chemical equation = C₆H₆(g) --> 6C---H(g)
Hence we can conclude that The answers to your questions are 5535.1 KJ/mol and C₆H₆(g) --> 6C---H(g)
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