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

A

+524.1

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B

+41.2

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C

-262.5

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D

-41.2

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Solution

The correct option is B
+41.2

(i) $C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J m o l^{- 1}$

(ii) $C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H = - 110.5 k J m o l^{- 1}$

(iii) $H_{2} (s) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 241.8 k J m o l^{- 1}$

For getting $C O_{2} (g) + H_{2} (g) \to C O (g) + H_{2} O (g),$ add

(ii) and (iii) and subtract (i).

Thus $- 110.5 - 241.8 + 393.5 = 395.3$

= $41.2 k J m o l^{- 1}$

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Similar questions

Δ H_{f}^{\circ}

C O_{2} (g)

C O (g)

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

Δ_{f} H^{0}

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)

Δ H_{f}^{o}

C O_{2 (g)}

C O_{(g)}

H_{2} O_{(g)}

- 393.5

- 110.5

- 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)}

Δ H_{t}^{\circ}

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

H_{2} O_{(g)}

- 393.5, - 110.5

- 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)}

Δ_{f} H^{\circ}

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

- 393.5, - 110.5

- 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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