Calculate the enthalpy change for the reaction (only magnitude in nearest integer in kj/mol):
$2 C (s) + 2 H_{2} (g) + O_{2} (g) ⟶ {C H}_{3} {C O}_{2} H (l)$

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Solution

Given reaction
$R : 2 C + 2 H_{2} + O_{2} \to C H_{3} C O_{2} H; Δ H = ?$

From the given table, enthalpies of the following reactions are

$R_{1} : C + O_{2} \to C O_{2}; Δ H_{1} = - 394 k J / m o l$

$R_{2} : H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H_{2} = - 286 k J / m o l$

$R_{3} : C H_{3} C O_{2} H + 2 O_{2} \to 2 C O_{2} + 2 H_{2} O; Δ H_{3} = - 876 k J / m o l$

And

$R = - R_{3} + 2 R_{1} + 2 R_{2}$

$\Rightarrow Δ H = - Δ H_{3} + 2 Δ H_{1} + 2 Δ H_{2} = + 876 + 2 \times (- 394) + 2 \times (- 286) = - 484 k J / m o l$

Answer = 484

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C_{2} H_{4}

C_{2} H_{6}

C_{2} H_{4} (g) + H_{2} (g) ⟶ C_{2} H_{6} (g)

Bond	$C - C$	$C = C$	$C - H$	$H - H$
Bond Enthalpy	$336.81 k J / m o l$	$606.68 k J / m o l$	$410.87 k J / m o l$	$431.79 k J / m o l$

C_{2} H_{4}

C_{4} H_{6}

α

C_{12} H_{22} O_{11} (a q) + H_{2} O (l) ⟶ 2 C_{6} H_{12} O_{6} (a q)

Δ_{f} H^{o} (H_{2} O, l) = - 285.85 k J

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Δ_{f} H^{o} (C_{12} H_{22} O_{11}, a q) = - 2238.3.1 k J

{m o l}^{- 1}

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