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Heat of Reaction

The reaction ...

Question

The reaction of heated graphite with superheated steam is endothermic;
$C (g r a p h i t e) + H_{2} O (g) \to C O g) + H_{2} (g) 1$
The heat required for this reaction is provided by burning part of the graphite;
$C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); △ H^{\circ} (500 K) = - 393.7 k J m o l^{- 1} 2$ .
If 1.34 mole of graphite is required for production of 1 mole of Hydrogen at 500 K.
What is $△ H^{\circ} (500 K)$ for reaction (1) ?

134 k J m o l^{- 1}

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144 k J m o l^{- 1}

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234 k J m o l^{- 1}

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244 k J m o l^{- 1}

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Solution

The correct option is A $134 k J m o l^{- 1}$
$C (g) + H_{2} O (g) \to C O (g) + H_{2} (g)$
heat required is provided by
$C (g) + O_{2} (g) \to C O_{2} (g)$
$△ H^{\circ} (500 K) = - 393.7 k J m o l^{- 1}$
1.34 mole of graphite is required for production of 1 mole of hydrogen. To produce 1 mole hydrogen, 1 mole graphite will be required in first reaction.
This means that 0.34 mole extra substance is used for production of heat
$∴ △ H^{\circ}$ for reaction $\Rightarrow 0.34 \times 393.7 = 133.85 k J m o l^{- 1} \approx 134$ $k J m o l^{- 1}$

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Similar questions

C (g r a p h i t e) + H_{2} O (g) \to C O g) + H_{2} (g) 1

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); △ H^{\circ} (500 K) = - 393.7 k J m o l^{- 1} 2

△ H^{\circ} (500 K)

C (g r a p h i t e) + H_{2} O (g) \to C O g) + H_{2} (g) 1

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); △ H^{\circ} (500 K) = - 393.7 k J m o l^{- 1} 2

△ H^{\circ} (500 K)

[i . e ., C_{g r a p h i t e} + H_{2} O (g) ⟶ C O (g) + H_{2} (g)]

O_{2} (g)

[C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 393.7 k J m o l^{- 1} a t 500 K]

H_{2}

H_{2} O

+ O_{2} \to C O_{2} (g); Δ H = - 94.3 k c a l

+ O_{2} \to C O_{2} (g); Δ H = - 97.6 k c a l

\to

C (d i a m o n d, s) + O_{2} (g) \to C O_{2} (g); △ H = - 94.3 k c a l

C (g r a p h i t e, s) + O_{2} (g) \to C O_{2} (g); △ H = - 97.6 k c a l

C (d i a m o n d, s) \to C (g r a p h i t e, s)

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