Calculate the enthalpy of formation of ethene $C_{2} H_{4}$ (only magnitude in nearest integer in kj/mol).

Open in App

Solution

Formation of $C_{2} H_{4}$

$R : 2 C + 2 H_{2} \to C_{2} H_{4}; Δ H = ?$

From the given table, enthalpies of the following reactions are
$R_{1} : C + O_{2} \to C O_{2}; Δ H_{1} = - 394 k J / m o l$

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

$R_{6} : C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O; Δ H_{3} = - 1393 k J / m o l$

And

$R = - R_{6} + 2 R_{1} + 2 R_{2}$

$\Rightarrow Δ H = - Δ H_{6} + 2 Δ H_{1} + 2 Δ H_{2} = 1393 - 2 \times 394 - 2 \times 286 = 33 k J / m o l$

Answer = 33

Suggest Corrections

Similar questions

C_{2} H_{4}

C_{2} H_{6}

C_{4} 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	$336.81 k J / m o l$	$606.68 k J / m o l$	$410.87 k J / m o l$	$431.79 k J / m o l$

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

S - S

(i) C_{2} H_{5} - S - C_{2} H_{5} (g)

Δ H_{f}^{o} = - 147.2 k J / m o l

(i i) C_{2} H_{5} - S - S - C_{2} H_{5} (g)

Δ H_{f}^{o} = - 201.9 k J / m o l

(i i i) S (g)

Δ H_{f}^{o} = 222.8 k J / m o l

Join BYJU'S Learning Program