-f H- - CO - -c H- graphite - - c H- COIf true 1 else 0-

The correct option is A 1As we know, C(graphite) + O_2(g) → CO_2(g) nbsp;H = 393.5 kJ ................1 C(graphite) + 1/2 O_2(g) → CO(g) nbsp; ...

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Byju's Answer

Standard XII

Physics

Banking of Roads

Δf H⊖ , CO ...

Question

$Δ_{f} H^{⊖}$ , $C O$ = $Δ_{c} H^{⊖} (g r a p h i t e) - Δ_{c} H^{⊖} (C O)$
If true 1 else 0.
1

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Solution

The correct option is A 1
As we know,
$C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g) H = - 393.5 k J$ ................1
$C (g r a p h i t e) + 1 / 2 O_{2} (g) \to C O (g)$ $Δ_{f} H^{⊖}$ $C O$ .......................2
and
$C O + 1 / 2 O_{2} (g) \to C O_{2} (g)$ $Δ_{c} H^{⊖}$ $C O$ .......................3
From these equations, it can be seen that
$Δ_{f} H^{⊖}$ $C O$ = $Δ_{c} H^{⊖} (g r a p h i t e) - Δ_{c} H^{⊖} (C O)$

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Similar questions

Q. Which of the following has the same value as

Δ_{f} H^{⊖}

C O

\frac{1}{2} Δ_{f} H^{⊖} (C O_{2})

\frac{1}{2} Δ_{c} H^{⊖} (g r a p h i t e)

Δ_{f} H^{⊖} (C O_{2}) - Δ_{f} H^{⊖} (g r a p h i t e)

Δ_{c} H^{⊖} (g r a p h i t e) - Δ_{c} H^{⊖} (C O) ?

{C H}_{4 (g)} + 1 / 2 O_{2 (g)} ⟶ {C H}_{3} {O H}_{(l)}

Δ_{c} H

{C H}_{4} = - x

and

Δ_{c} H

{C H}_{3} O H = - y

Then,

Δ H

for reaction:

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
(a) $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{2} O (l); Δ_{c} H = - 2658.0 k J m o l^{- 1}$
(b) $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{c} H = - 1329.0 k J m o l^{- 1}$
(c) $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{c} H = - 2658.0 k J m o l^{- 1}$
(d) $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l); Δ_{c} H = + 2658.0 k J m o l^{- 1}$

Q. At

298 K

Δ_{f} H^{⊖} e l e m e n t = 0

.
If true 1 else 0

C H_{3} O H (1) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (I); Δ, H^{⊖} = - 726 k J m o l^{- 1}

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ_{c} H^{⊖} = - 393 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (I); Δ_{f} H^{⊖} = - 286 k J m o l^{- 1}

Calculate the standard enthalpy of formation of

{C H}_{3} O H (l)

from the following data:

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ΔfH⊖ , CO = ΔcH⊖(graphite)−ΔcH⊖(CO)If true 1 else 0.1

The correct option is A 1As we know,C(graphite)+O2(g)→CO2(g)H=−393.5kJ................1C(graphite)+1/2O2(g)→CO(g) ΔfH⊖ CO.......................2andCO+1/2O2(g)→CO2(g) ΔcH⊖ CO.......................3From these equations, it can be seen thatΔfH⊖ CO = ΔcH⊖(graphite)−ΔcH⊖(CO)

$Δ_{f} H^{⊖}$ , $C O$ = $Δ_{c} H^{⊖} (g r a p h i t e) - Δ_{c} H^{⊖} (C O)$
If true 1 else 0.
1