The reaction of heated graphite with superheated steam is endothermic-C graphite - H 2 O g - COg - H 2 g 1The heat required for this reaction is provided by burning part of the graphite- C - graphite - O - 2 - g rightarrow CO - 2 - g -If 1-34 mole of graphite is required for production of 1 mole of Hydrogen at 500 K- What is H -500 K for reaction 1-A- 134 kJmol -1B- 234 kJ mol -1C- 244 kJmol -1D- 144 kJ mol -1

The correct option is A 134 kJmol^ 1C (g) + H_2O (g) → CO (g) + H_2 (g) heat required is provided by nbsp; C (g) + O_2 (g) → CO_2 (g) H^∘ (500 K) = 393.7 ...

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Standard XII

Chemistry

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

Q. 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.
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△ H^{\circ} (500 K)

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Q. The endothermic reaction of carbon (graphite) in super heated steam at 500 K

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

is made by providing heat by burning a part of carbon (graphite) in

O_{2} (g)

followed by an exothermic reaction :

[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]

If a total of 1.34 mole of graphite is required for the production of 1 mole of

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Heat of Reaction

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