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Byju's Answer

Standard XII

Chemistry

Vant Hoff's Equation and Effect of Temperature on Equilibrium Constant

The Δ H f ∘ f...

Question

The $Δ H_{f}^{\circ}$ for $C O_{2} (g)$ , $C O (g)$ and $H_{2} O (g)$ are - 393.5, -110.5 and - 241.8 KJ $m o l^{- 1}$ respectively.
The standard enthalpy change (in kJ) for the reaction $C O_{2} + H_{2} \to C O + H_{2} O$ is:

524.1

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41.2

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-262.5

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-41.2

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Solution

The correct option is B 41.2
Given
$C + O_{2} \to C O_{2}$ ; $Δ H^{\circ} = - 393.5 k J$ ....(i)
$C + \frac{1}{2} O_{2} \to C O$ ; $Δ H^{\circ} = - 110.5 k J$ ....(ii)
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ ; $Δ H^{\circ} = - 241.8 k J$ ....(iii)
By Eqs. (ii) + (iii) - (i)
$C O_{2} + H_{2} \to C O + H_{2} O$ ; $Δ H^{\circ} = 41.2$

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Similar questions

The $Δ H_{f}^{\circ}$ for $C O_{2} (g), C O (g)$ and $H_{2} O (g)$ are $- 393.5, - 110.5$ and $- 241.8 k J m o l e^{- 1}$ 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. The

Δ_{f} H^{0}

for

C O_{2} (g), C O (g) a n d H_{2} O (g)

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 reaction

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

is :

Q. The

Δ H_{f}^{0}

for

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

and

H_{2} O (g)

are

- 393.5, - 110.5

and

- 241.8 k J 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)

Q. The

Δ H_{t}^{\circ}

for

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

and

H_{2} O_{(g)}

are

- 393.5, - 110.5

and

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

respectively the standard enthalpy change (in

K J

) for the reaction

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

is:

Q. The

Δ_{f} H^{\circ}

for

C O_{2} (g), C O (g) a n d H_{2} O (g)

are

- 393.5, - 110.5

and

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

respectively. The standard enthalpy change (in kJ) for the reaction is:

C O_{2} (g) + H_{2} (g) ⟶ C O (g) + H_{2} O (g)

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The ΔH∘f for CO2(g), CO(g) and H2O(g) are - 393.5, -110.5 and - 241.8 KJ mol−1 respectively. The standard enthalpy change (in kJ) for the reaction CO2+H2→CO+H2O is:

The correct option is B 41.2Given C+O2→CO2; ΔH∘=−393.5kJ ....(i) C+12O2→CO; ΔH∘=−110.5kJ ....(ii) H2+12O2→H2O; ΔH∘=−241.8kJ ....(iii) By Eqs. (ii) + (iii) - (i) CO2+H2→CO+H2O; ΔH∘=41.2

The $Δ H_{f}^{\circ}$ for $C O_{2} (g)$ , $C O (g)$ and $H_{2} O (g)$ are - 393.5, -110.5 and - 241.8 KJ $m o l^{- 1}$ respectively.
The standard enthalpy change (in kJ) for the reaction $C O_{2} + H_{2} \to C O + H_{2} O$ is: