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Hess' Law

If C+O2.⟶CO...

Question

If $C + O_{2} . ⟶ {C O}_{2} + 94.2 k c a l$
$H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} + 68.3 k c a l$
${C H}_{4} + 2 O_{2} ⟶ {C O}_{2} + 2 H_{2} O + 210.8 k c a l$
Then, the heat of formation of methane will be

47.3 k c a l

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20.0 k c a l

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45.9 k c a l

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- 47.3 k c a l

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Solution

The correct option is B $20.0 k c a l$
$C + O_{2} ⟶ {C O}_{2} + 94.3 k c a l . . . . . . . . . (i)$
$H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O + 68.3 k c a l . . . . . . . (i i)$
On multiplying Eq. $(i i)$ by $2$ and then adding in Eq. $(i)$ , we get
$C + 2 H_{2} + 2 O_{2} ⟶ {C O}_{2} + 2 H_{2} O + 230.8 k c a l . . . . . (i i i)$
On substracting, adding the following equation from Eq. $(i i i)$
${C H}_{4} + 2 O_{2} ⟶ {C O}_{2} + 2 H_{2} O + 210.8 k c a l$
we get
$C + 2 H_{2} ⟶ {C H}_{4} + 20 k c a l$

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Similar questions

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

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C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O

2 C_{2} H_{2} + 5 O_{2} ⟶ 4 C O_{2} + 2 H_{2} O

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O_{2}

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S + O_{2} \to {S O}_{2}; Δ H = - 298.2 k J

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H_{2} {S O}_{4}

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