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Spontaneity

0.5 H 2 g +0....

Question

$0.5 H_{2} (g) + 0.5 C l_{2} (g) \to H C l (g); Δ H_{298}^{0} = - 22060 c a l$

The mean heat capacities over this temperature range are,

$H_{2} (g), C_{p} = 6.82 c a l m o l^{- 1} K^{- 1}; C l_{2} (g), C_{p} = 7.7 c a l m o l^{- 1} K^{- 1}; H C l (g), C_{p} = 6.81 c a l m o l^{- 1} K^{- 1}$

The magnitude of the heat of formation of $H C l (g)$ (in cal) at $348 K$ from the above data will be:

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Solution

$Δ C_{p} = C_{p} (H C l) - 0.5 C_{p} (H_{2}) - 0.5 C_{p} (C l_{2})$
$= 6.81 - 3.41 - 3.85 = - 0.45$
$Δ H_{348} = Δ H_{298} + Δ C_{p} \times Δ T$
$Δ H_{348}$ = $- 22060 + (- 0.45) (50) = - 22082.5 c a l$

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Similar questions

H C l

\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g) △ H_{298}^{o} = - 22060 c a l m o l^{- 1}

H_{2} (g), C_{P} = 6.82 c a l m o l^{- 1} K^{- 1};

C l_{2} (g), C_{P} = 7.71 c a l m o l^{- 1} K^{- 1};

H C l (g), C_{P} = 6.81 c a l m o l^{- 1} K^{- 1} .

H C l

\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g) △ H_{298}^{o} = - 22060 c a l m o l^{- 1}

H_{2} (g), C_{P} = 6.82 c a l m o l^{- 1} K^{- 1};

C l_{2} (g), C_{P} = 7.71 c a l m o l^{- 1} K^{- 1};

H C l (g), C_{P} = 6.81 c a l m o l^{- 1} K^{- 1} .

C_{v} = 650 J K g^{- 1} K^{- 1}

C_{p} = 910 J K g^{- 1} K^{- 1}

H_{2} (g) + C l_{2} (g) \to 2 H C l (g)

Δ H

- 44

10^{\circ} C t o 20^{\circ} C .

C_{p} = 7.03

^{- 1}

^{- 1}

R = 8.31 J

^{- 1} K^{- 1}

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