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Shape and Geometry Using VSEPR

Given that: ...

Question

Given that:

$C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 393.7 k J$

Calculate the quantity of graphite (in gm) that must be burnt to evolve $5000 k J$ of heat.

152.8

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15.2

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200

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None of these

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Solution

The correct option is A 152.8
As given,
$C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 393.7 k J$

As one mole of graphite means 12 gm gives 393.7 kJ of heat.

Therefore, 5000 kJ of heat will be given by $= \frac{12 \times 5000}{393.7} = 152.8 g m$ graphite.

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Similar questions

C_{(g r a p h i t e)} + O_{2 (g)} \to C O_{2 (g)}; Δ H = - 393.7 k J

C_{(d i a m o n d)} \to C_{(g r a p h i t e)}; Δ H = - 2.1 k J

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.

Δ H

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

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 393.5 k J

C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 395.4 k J

Δ H

C (d i a m o n d) + O_{2} \to C O_{2} (g); Δ H = - 94.5 K c a l

C (g r a p h i t e) + O_{2} \to C O_{2} (g); Δ H = - 94 K c a l

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