Question

2. Given the enthalpy of hydrogenation of benzene (CeHs reacts with Ha forming CoH) (A,H= +215 kl/mol), the enthalpy of combustion of benzene (CeHe reacts with Oz forming CO2 and H O) (4 H = -3268 kl/mol 1 and the reaction enthalpy of water decomposition (liquid H,O becomes H2 and O2 gases) (A,H = +286 k]/mol), what is the enthalpy of combustion of cyclohexane (A H°). (hint: write all balanced equations, use Hess' Law) (15 pts)

Answer 1

According to the Lavoisier and Laplace :

The heat changes for a reaction in forward direction is equal in the magnitude but opposite in sign for the reverse reaction.

If A. ---> B . ∆H = x KJ

For reverse reaction

B ---> C . ∆H = - x KJ

According to the Hess's law :

For a chemical reaction , the net heat change (∆H) will be same weather the process occurs in one step or in several step.

Let a chemical reaction occurs in one step

A --- > C . ∆H =∆ H _{​​​​​1}

But when the same reaction occurs in two step

A ---> B . ∆H =∆H_{​​​​​​2}

B ---> C . ∆H = ∆H₃

Then according to the law

∆H₁ = ∆H₂ + ∆H₃

Given that C6H6 + H₂ = Colt I H = +215 kJ/mol Có H t 2 = cot tro & Hºc - 3368 kymax H2O = H₂(g) + Q iG) AH = + 286 kJ/mol we have to determine combustion of Cyclonexone, . Cotia to = c + H₂O

Balancing the equation CHO + 3H2 = Catto AH = +215 Kalmar CHE + 150; = 6CO2 + 3H2O TH°= -326367lmek H2O = H2 + 20 Kg HP = + 286 kJ/mol we have to determine: TCHF 4H : 6 Hip t 902 = 6CO₂ + 6H20

C6H12. = C to + 3H2 ApH = -215 kJ/mol G Hi te g = 6ą. +340 VH = -3768kJ/mol (H₂ + 1/2O2 = H2O Pothi=-268 kJ/mod) x3 Hence we get . Cohip = G H + 3H₂ T j H = -215kJ/mal & H + 0, = 6co, + 3H2O (H° =-3268 kg/cme 3H2 + 0 = 3H20 ritto -804 kJ/mol Co Hią + 9 = 6 CQ +6H2O DH = VyHº+keHº+KpH I - 4H =(-345-3369-864)<3/nax 3-4287 kJ/mol Hence the combustion enthalpy of Cyciohexane is -4287. Kolmal.