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 $H_{2}$ at 500 K, the heat of reaction for carbon (graphite) with $H_{2} O$ in kJ is ________.

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Solution

One mole of $H_{2}$ will be formed by 1 mole graphite following the reaction:
$C (g r a p h i t e) + H_{2} O (g) ⟶ C O (g) + H_{2} (g); Δ H^{\circ} = ?$ ... (i)
Thus, mole of graphite used by the reaction (ii)
$= (1.34 - 1) = 0.34$
$C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H^{\circ} = - 393.7 k J$ ... (ii)
$T h e r e f o r e, h e a t p r o d u c e d i n (i i) b y 0.34 m o l e C (g r a p h i t e) = h e a t a b s o r b e d i n (i) b y 1 m o l e C (g r a p h i t e)$
$0.34 \times Δ H_{(i i)} = 1 \times Δ H_{(i)}$
or $0.34 \times 393.7 = 1 \times Δ H_{(i)}$
$∴ Δ H_{(i)} = 133.86 k J$

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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)

Δ H_{t}

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

Δ H = x k J

C (d i a m o n d) + O_{2} (g) \to C O_{2} (g)

Δ H = y k J

Assertion- $C (g r a p h i t e) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 393.5 k . J$

$C (d i a m o n d) + O_{2 (g)} \to C O_{2 (g)}$ ,

$Δ H = - 395.4 k . J$

Reason(R): graphite is more stable than diamond, so heat of combustion is less.

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g) : H = - x_{1} c a l

C (d i a m o n d) + O_{2} (g) \to C O_{2} (g) : H = - x_{2} c a l

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