The -heat of total cracking- of hydrocarbons - H TC is defined as - H at 298-15 K and 101-325 kPa for the process below- C n H m - 2n- m 2 H 2 g- n CH 4 gGiven that- - H TC is -65-2 kJ for C2H6 and -87-4 kJ for C3H8- calculate - H for CH 4 g- C 3 H 8 g- 2 C 2 H 6 g

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Physical Properties of Aromatic Compounds

The "heat of ...

Question

The "heat of total cracking" of hydrocarbons $Δ H_{T C}$ is defined as $Δ H$ at $298.15 K$ and $101.325 k P a$ for the process below:
$C_{n} H_{m} + (2 n - \frac{m}{2}) H_{2} (g) ⟶ n {C H}_{4} (g)$
Given that, $Δ H_{T C}$ is $- 65.2 k J$ for $C_{2} H_{6}$ and $- 87.4 k J$ for $C_{3} H_{8}$ , calculate $Δ H$ for
${C H}_{4} (g) + C_{3} H_{8} (g) ⟶ 2 C_{2} H_{6} (g)$

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C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {C O}_{2} (g) + 4 H_{2} O (l)

Δ H - Δ U

C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)

Δ H - Δ E

Δ H - Δ U

C_{3} H_{8}

{5 O}_{2}

⟶

{3 C O}_{2}

{4 H}_{2} O (ℓ)

25^{o} C

1

{C H}_{4} (g) ⟶ C (g) + 4 H (g)

Δ_{r} H = 396 k c a l / m o l e

C_{2} H_{6} (g) ⟶ 2 C (g) + 6 H (g)

Δ_{r} H = 676 k c a l / m o l e

C - C

C_{2} H_{6}

C_{2} H_{6} (g)

Δ_{s u b} C (s) = 171.8 k c a l / m o l e

B . E (H - H) = 104.1 k c a l / m o l e

C (d i a m o n d) + 2 H_{2} (g) ⟶ C H_{4} (g)

Δ H

Δ U

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