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

Byju's Answer

Standard XII

Chemistry

Heat of Reaction

COg + 12 O2g ...

Question

$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$
The value of standard change in enthalpy for the above reaction on the basis of the following data is:
$T = 300 K, Δ S^{o} = - 0.094 k J m o l^{- 1} K^{- 1}, Δ G_{f}^{o} (C O_{2}) = - 394.4 k J m o l^{- 1}$ and $Δ G_{f}^{0} (C O) = - 137 k J m o l^{- 1}$

- 285.4 k J / m o l

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

- 28.54 k J / m o l

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

285.4 k J / m o l

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 A $- 285.4 k J / m o l$
$C O + \frac{1}{2} O_{2} ⇌ C O_{2}$
$Δ G_{r K}^{o} = Δ G_{f}^{o} (C O_{2}) - Δ_{f} G^{o} (C O)$
$= - 394.4 + 137$
$= - 257.4 k J / m o l$
$∴ Δ H_{r k n} - T Δ S = - 257.4$
$\Rightarrow Δ H = - 257.4 + 300 \times (- 0.094)$
$= - 257.4 + (- 28.2)$
$= - 285.6 k J / m o l$ .

Suggest Corrections

Similar questions

Q. Show that the reaction,

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g)

300 K

is spontaneous and exothermic, when the standard entropy change is

- 0.094 k J {m o l}^{- 1} K^{- 1}

. The standard Gibbs free energies of formation for

{C O}_{2}

and

C O

are

- 394.4

and

- 137 k J {m o l}^{- 1}

respectively.

Q. The

Δ_{f} H^{o}

for the

C O_{2} (g), C O (g)

and

H_{2} O (l)

are -393.5, -110.5 and -241.8

k J m o l^{- 1}

respectively. The standard enthalpy change (in

k J m o l^{- 1}

) for the given reaction is :

C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (l)

Q. The

Δ H_{f}^{o}

for

C O_{2 (g)}

C O_{(g)}

and

H_{2} O_{(g)}

are

- 393.5

- 110.5

and

- 241.8

m o l^{- 1}

respectively the standard enthalpy change (in KJ) for the reaction

C O_{2 (g)} + H_{2 (g)} \to C O_{(g)} + H_{2} O_{(g)}

is:

Q. Show that the reaction

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g)

300 K

is spontaneous and exothermic, when the standard entropy change is

- 0.094 k J m o l^{- 1} K^{- 1}

. The values of standard Gibbs free energy of formation of

C O_{2}

and

C O

are

- 394.4

and

- 137.2 k J m o l^{- 1}

, respectively.

Standard enthalpy of combustion of $C H_{4}$ is $- 890 k J m o l^{- 1}$ and standard enthalpy of vaporisation of water is 40.5 kJ $m o l^{- 1}$ . Calculate the enthalpy change for the reaction :
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$

Join BYJU'S Learning Program

CO(g)+12O2(g)→CO2(g)The value of standard change in enthalpy for the above reaction on the basis of the following data is:T=300 K, ΔSo=−0.094 kJmol−1K−1,ΔGof(CO2)=−394.4 kJmol−1 and ΔG0f(CO)=−137 kJmol−1

The correct option is A −285.4 kJ/molCO+12O2⇌CO2ΔGorK=ΔGof(CO2)−ΔfGo(CO)=−394.4+137=−257.4kJ/mol∴ΔHrkn−TΔS=−257.4⇒ΔH=−257.4+300×(−0.094) =−257.4+(−28.2) =−285.6kJ/mol.