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Heat of Formation

The enthalpy ...

Question

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$

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Solution

According to the question,

$C H_{4} (g) + 2 O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (l)$ , $Δ H = - 893 k J m o l^{- 1}$ .....(i)
$C (s) + O_{2} (g) ⟶ C O_{2} (g)$ , $Δ H = - 393.5 k J m o l^{- 1}$ .........(ii)
$H_{2} + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l)$ , $Δ H = - 285.8 k J m o l^{- 1}$ .......(iii)
$C (s) + 2 H_{2} (g) ⟶ C H_{4} (g)$ , $Δ H = ?$ ......(iv)
Equation (ii)+2× Equation (iii) -Equation (i) gives the required equation (iv) with $Δ H$ = −393.5+2(−285.8)−(−890.3) $k J m o l^{- 1}$ = -74.8 $k J m o l^{- 1}$

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Similar questions

298 K

- 890.3

k J {m o l}^{- 1}, - 393.5

k J {m o l}^{- 1}, - 285.8

k J {m o l}^{- 1}

{C H}_{4 (g)}

298 K

- 890.3

m o l^{- 1}

- 285.8

m o l^{- 1}

C H_{4} (g)

C_{(g r a p h i t e)} + O_{2} (g) \to C O_{2} (g);

Δ_{f} H^{\circ} = - 393.5 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

Δ_{f} H^{\circ} = - 285.8 k J m o l^{- 1}

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

Δ_{f} H^{\circ} = + 890.3 k J m o l^{- 1}

Δ_{f} H^{\circ}

C_{(g r a p h i t e)} + 2 H_{2} (g) \to C H_{4} (g)

{C H}_{4} (g) + 2 O_{2} (g) ⟶ {C O}_{2} (g) + 2 H_{2} O (l)

{C H}_{4}, {C O}_{2}

H_{2} O

74.8 k J {m o l}^{- 1}, - 393.5 k J {m o l}^{- 1}, - 286 k J {m o l}^{- 1}

- 890.3 k J^{- 1}

- 393.5 k J^{- 1}

- 285.8 k J^{- 1}

{C H}_{4} (g)

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