For reaction C 2 H 4 g-3 O 2 g- 2C O 2 g-2 H 2 Ol- U-1415kJ- Then - H at 27oC is -

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Heat of Reaction

For reaction ...

Question

For reaction
$C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (l); Δ U = - 1415 k J$ . Then $Δ H$ at $27^{o} C$ is :

- 1410 k J

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- 1420 k J

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+ 1420 k J

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+ 1410 k J

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Solution

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Similar questions

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)

Δ E = - 1415 K J

Δ H

27 C

:For the reaction

$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l);$

$Δ$ $E$ = $- 1415 k J$ . The $Δ$ H at 27 $^{0}$ C is

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 {C O}_{2} (g) + 2 H_{2} O (g)

C_{2} H_{2}

C_{2} H_{4}

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

C_{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ 2 C O_{2} (g) + H_{2} O (l); Δ H = - 310.61 k c a l

C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + H_{2} O (l); Δ H = - 337.32 k c a l

C_{2} H_{4} (g) + 3 O_{2} \to 2 C O_{2} (g) + 2 H_{2} O (l)

Δ U = - 1415

Δ H

27^{o}

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