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Heat of Formation

The heat of t...

Question

The heat of transition ( $Δ H_{t}$ ) of graphite into diamond would be,
where, $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$ $Δ H = x k J$
$C (d i a m o n d) + O_{2} (g) \to C O_{2} (g)$ $Δ H = y k J$

A

x + y

kJ/mole

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B

x - y

kJ/mole

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C

y - x

kJ/mole

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D

None of these

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Solution

The correct option is B $x - y$ kJ/mole
To calculate:
$C (g r a p h i t e) \to C (d i a m o n d)$
Heat of reaction will be equal to the heat of transition
Given:
$C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$ $Δ H = x k J$
$C O_{2} (g) \to C (d i a m o n d) + O_{2} (g)$ $Δ H = - y k J$
So, by adding the two, required $Δ H$ = $(x - y$ ) kJ/mole

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Similar questions

Q. The heat of transition (

Δ H_{t}

) of graphite into diamond would be,
where,

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

Q. Consider the following reaction :

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

What is the heat of transition of graphite into diamond?

Q. Determine the heat of transformation of C(diamond)→ C(graphite) from the following data.

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

C_{(g r a p h i t e)} + O_{2} (g) \to C O_{2} (g) △ H = - 94.0 K c a l

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

$Δ H = - 94.05 k c a l m o l^{- 1}$

$C_{d i a m o n d} + O_{2} (g) \to C O_{2} (g); Δ H = - 94.50 k c a l m o l^{- 1}$ therefore

Q. One gram mole of graphite and diamond were burnt to form

{C O}_{2}

C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 399.5 k J

C_{d i a m o n d} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 395.4 k J

then, which of the following is the correct about diamond and graphite?

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Heat of Formation

Standard XII Chemistry

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