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Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

The Δ H f ∘ f...

Question

The $Δ H_{f}^{\circ}$ for $C O_{2} (g)$ , CO(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
$Δ_{r} H^{\circ} = Δ_{f} H^{\circ} (C O) + Δ_{f} H^{\circ} (H_{2} O) - Δ_{f} H^{\circ} (H_{2})$
=-110.5-241.8+393.5-0
=41.2

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Similar questions

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 :

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 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)

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)

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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 41.2ΔrH∘=ΔfH∘(CO)+ΔfH∘(H2O)−ΔfH∘(H2) =-110.5-241.8+393.5-0 =41.2

The $Δ H_{f}^{\circ}$ for $C O_{2} (g)$ , CO(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:

The correct option is B 41.2
$Δ_{r} H^{\circ} = Δ_{f} H^{\circ} (C O) + Δ_{f} H^{\circ} (H_{2} O) - Δ_{f} H^{\circ} (H_{2})$
=-110.5-241.8+393.5-0
=41.2