Calculate the heat of combustion of benzene from the following data-a- 6Cs - 3H2g- C6H6l- - H - 49-0 kJb- H2g - 1-2O2g- H2Ol- - H - -285-8 kJc- Cs - O2g- CO2g- - H - -389-3 kJ

The correct option is C Δ_c H = 3242.2 kJ mol^ 1 As given,a. #160; 6C(s) + 3H_2(g)⟶ C_6H_6(l), Δ H_1 = 49.0 kJ b. #160; H_2(g) + 1/2O_2(g)⟶ ...

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Enthalpy of Combustion

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Question

Calculate the heat of combustion of benzene from the following data:

a. $6 C (s) + 3 H_{2} (g) ⟶ C_{6} H_{6} (l), Δ H = 49.0 k J$

b. $H_{2} (g) + 1 / 2 O_{2} (g) ⟶ H_{2} O (l), Δ H = - 285.8 k J$

c. $C (s) + O_{2} (g) ⟶ C O_{2} (g), Δ H = - 389.3 k J$

Δ_{c} H = - 3242.2 k J m o l^{- 1}

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Δ_{c} H = + 3242.2 k J m o l^{- 1}

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Δ_{c} H = - 4242.2 k J m o l^{- 1}

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None of these

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6 C (s) + 3 H_{2} (g) \to C_{6} H_{6} (l); Δ H = + 45.9 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 285.9 k J

C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J

C (s) + O_{2} (g) ⟶ C O_{2} (g)

Δ H = - 393.7 k J

H_{2} (g) + 1 / 2 O_{2} (g) ⟶ H_{2} O (l)

Δ H = - 285.8 k J

C H_{3} C O O H (l) + 2 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (l)

Δ H = - 873.2 k J

{C H}_{3} O H (l)

{C H}_{3} O H (l) + \frac{3}{2} O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (l); Δ H = - 726 k J

C (s) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 394 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 k J

C H_{3} O H (l)

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l); △_{r} H^{⊖} = - 726 k J

C (s) + O_{2} (g) \to C O_{2} (g), Δ H = - 394 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H = - 286 k J

C H_{3} C O O H (l)

25^{\circ} C

C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} + 2 H_{2} O (l); Δ H = - 208.34

C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 94.05

H_{2} (g) + \frac{1}{2} O_{2} \to H_{2} O (l); Δ H = - 68.32

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