The heat of transition - H t of graphite into diamond would be- where- C graphite - O 2 g - CO 2 g - H - x kJC diamond - O 2 g - CO 2 g - H - y kJA- x - y kJ - moleB- x - y kJ - moleC- y - x kJ - moleD- None of these

The correct option is B x y kJ/moleTo calculate: nbsp; C (graphite) → nbsp;C (diamond) Heat of reaction will be equal to the heat of transition Given: C ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Heat of Formation

The heat of t...

Question

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

x + y

kJ/mole

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x - y

kJ/mole

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y - x

kJ/mole

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

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}

gas.

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?

Join BYJU'S Learning Program

Explore more

Heat of Formation

Standard XII Chemistry

Join BYJU'S Learning Program

The heat of transition (ΔHt) of graphite into diamond would be, where, C(graphite)+O2(g)→CO2(g) ΔH=x kJ C(diamond)+O2(g)→CO2(g) ΔH=y kJ

The correct option is B x−y kJ/moleTo calculate: C(graphite)→C(diamond) Heat of reaction will be equal to the heat of transition Given: C(graphite)+O2(g)→CO2(g) ΔH=x kJ CO2(g)→C(diamond)+O2(g) ΔH=−y kJ So, by adding the two, required ΔH = (x−y) kJ/mole

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