Question

(2). Consider benzene in the gas phase, C&H(g). Use the heat of formation, AH, values below to answer the questions. (20 points дня (кJ/mo) Substance H(g) C(g) 217.94 718.4 82.9 (8°HӘ (a) What is the standard enthalpy change for the reaction that represents breaking all the bonds in gaseous benzene, CHg)? (6 points) (b) What is the chemical equation for the reaction that corresponds to breaking just the carbon-carbon bonds in gaseous benzene, CeH6()? Indicate the phase of each species in your answer. (4 points) (c) 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: CHI() — C(g) + Hig), ДН3413 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

Given

1/2 H₂(g) -----> H(g) $\Delta$H^o_f = 217.94 kJ/mol ------(1)

C(s) ------> C(g) $\Delta$H^o_f = 718.4 kJ/mol ------(2)

6C(s) + 3H₂(g) ----> C₆H₆ (g) $\Delta$H^o_f = 82.9 kJ/mol ------(3)  

Required to find:

C₆H₆ (g) ------> 6C(g) + 6H(g)   

Multiplying eqn(1) by 6, eqn (2) by 6 and adding we get,

6C(s) + 3H₂(g) ------> 6C(g) + 6H(g) $\Delta$H = 6*217.94 + 6* 718.4 = 5618.04 kJ/mol ----(4)

Subtracting eqn (3) from eqn (4) we get,

6C(s) + 3H₂(g) - 6C(s) - 3H₂(g) ----> 6C(g) + 6H(g) - C₆H₆ (g)   $\Delta$H = 5618.04 - 82.9

=> C₆H₆ (g) ------> 6C(g) + 6H(g)    $\Delta$H = 5535.14 kJ/mol ----(5)

Also given,

C-H(g) -----> C(g) + H(g)   $\Delta$H = 413 kJ/mol ----(6)

Multiplying eqn (6) by 6 we get

6C-H(g) ------> 6C(g) + 6H(g)   $\Delta$H = 2478 kJ/mol ------(7)

Subtracting eqn (7) from eqn (5) we get,

C₆H₆ (g) - 6C-H(g) ----> 6C(g) + 6H(g) - 6C(g) - 6H(g) $\Delta$H = 3057.14 kJ/mol

=> C₆H₆ (g) ------> 6C-H(g)   $\Delta$H = 3057.14 kJ/mol

Therefore,

Standard enthalpy of breaking all bonds in C₆H₆(g) is 5535.14 kJ/mol

Standard enthalpy of breaking only C-C bonds in C₆H₆(g) is 3057.14 kJ/mol