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Standard XII

Chemistry

Le Chatelier's Principle for Del N Equals to 0

The Δ Hf0 f...

Question

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
$C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (g)$ 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 $41.2$

$△ H = H_{f}^{\circ} (p r o d u c t s) - H_{f}^{\circ} (r e a c t a n t s)$
Reactants are: $C O_{2} H_{2} a n d p r o d u c t s a r e C O a n d H_{2} O$
Hence,
$Δ H$ $= - 100.5 + (- 241.8) - (- 393.5 + 0) = 41.2 k J$
Hence, option B is correct

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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 :

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 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:

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^{\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 ΔH0f 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 reactionCO2(g)+H2(g)→CO(g)+H2O(g) is

The correct option is B 41.2△H=H∘f(products)−H∘f(reactants)Reactants are: CO2 H2 and products are CO and H2OHence,ΔH =−100.5+(−241.8)−(−393.5+0)=41.2 kJHence, option B is correct

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

The correct option is B $41.2$

$△ H = H_{f}^{\circ} (p r o d u c t s) - H_{f}^{\circ} (r e a c t a n t s)$
Reactants are: $C O_{2} H_{2} a n d p r o d u c t s a r e C O a n d H_{2} O$
Hence,
$Δ H$ $= - 100.5 + (- 241.8) - (- 393.5 + 0) = 41.2 k J$
Hence, option B is correct