Given- Cs-O2g-CO2g-94-2kcal H2g-12O2g-H2Ol-68-3kcal CH4-2O2g-CO2g-2H2Ol-210-8kcalThe heat of formation of methane in kcal will be-

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Hess' Law

Given: Cs+O2...

Question

Given:
$C (s) + O_{2} (g) \to {C O}_{2} (g) + 94.2 k c a l$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) + 68.3 k c a l$
${C H}_{4} + 2 O_{2} (g) \to {C O}_{2} (g) + 2 H_{2} O (l) + 210.8 k c a l$
The heat of formation of methane in $k c a l$ will be:

45.9

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47.8

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20.0

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47.3

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Solution

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C (s) + O_{2} (g) \to C O_{2} (g) Δ H = + 94.2 k c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) + 68.3 Δ H = k c a l

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) Δ H = + 210.8 k c a l

k c a l

C_{(s)} + O_{2 (g)} \to C O_{2 (g)};

Δ H = - 394 k J

2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)};

Δ H = - 568 k J

C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)};

Δ H = - 394 k J

C H_{4}

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

C (s) + O_{2} (g) \to C O_{2} (g) Δ H = - 94.2 k c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 68.3 k c a l

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) Δ H = - 210.8 k c a l

Heat of formation of $C H_{4}$ is:

If given:

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

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) △ H = - 284 k J$

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) △ H = - 892 k J$

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