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Calorimetry

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Question

Calculate the standard internal energy change for the reaction
$C_{(g r a p h i t e)} + H_{2} O (g) \to C O (g) + H_{2} O (g)$
$Δ_{f}^{o}$ at $25^{o}$ ; $H_{2} O (g) = - 241.8 k J {m o l}^{- 1}$
$C O (g) = - 110.5 k J {m o l}^{- 1}$
$R = 8.314 J K^{- 1} {m o l}^{- 1}$

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Solution

$\begin{matrix} Δ H = Δ U + Δ n g R T & Δ H = Δ H_{f} C O + Δ H_{f} H_{2} O Δ U = Δ H - Δ n g R T & \Rightarrow - 241.8 - 110.5 \Rightarrow - 352.3 - (2 - 1) 8.314 \times 298 & \Rightarrow - 352.3 K J \Rightarrow - 352.3 K J - (8.314 \times 298) J \Rightarrow - 352.3 K J - 2.477 K J \Rightarrow - 354.78 K J = Δ U \end{matrix}$

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Q. The equilibrium constant for the reaction

{C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

298 K

7

. Calculate the value of standard free energy change.

(R = 8.314 J K^{- 1} {m o l}^{- 1})

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 H_{f}^{0}

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 kJ) for the reaction

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

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)}

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