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Heat of Reaction

Given, H2Sg→ ...

Question

Given, $H_{2} S_{(g)} \to H S (g) + H (g); △ H_{r x n} = 379.6 k J m o l^{- 1}$
$H_{2} (g) + S (s) \to H_{2} S (g);$
$△ H_{f}^{\circ} H_{2} S = - 21.1$ $k J m o l^{- 1}$
$H_{2 (g)} \to 2 H_{(g)}$ ; $Δ H = 438 k J m o l^{- 1}$
$Δ H_{s u b l i m a t i o n}$ of S = $280 k J m o l^{- 1}$

Thus, $△ H_{f}^{\circ} H S$ is

A

138.5 k J m o l^{- 1}

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B

- 138.5 k J m o l^{- 1}

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C

- 277.0 k J m o l^{- 1}

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D

+ 139.5 k J m o l^{- 1}

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Solution

The correct option is D $+ 139.5 k J m o l^{- 1}$
$\frac{1}{2} H_{2} (g) + S (s) \to H S (g)$
This requires dissociation energy of $(H - S)$ bond of $H S (g)$ This can be derived from $△_{f} H^{\circ} (H_{2} S)$ .
$△_{f} H_{2} S (g)$ = [ $B E o f H_{2}$ + sublimation energy of $S (s)$ ] - [BE of first $H S$ bond + BE of second $H S$ bond]
$- 21.1 = [438 + 280] - [379.6 + x] ∴ x = 438 + 280 - 379.6 + 21.1 = 359.5 k J m o l^{- 1}$
For the formation of $H S (s)$
$\frac{1}{2} H_{2} (g) + S (s) \to H S (g)$
$△_{f} H (H S) = \frac{1}{2} B E (H_{2}) +$ sublimation energy of $S (s)$ - BE (second $H - S$ bond)
$= (219 + 280) - (359.5)$
$= 139.5$ $k J m o l^{- 1}$

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Similar questions

H_{2} S_{(g)} \to H S (g) + H (g); △ H_{r x n} = 379.6 k J m o l^{- 1}

H_{2} (g) + S (s) \to H_{2} S (g);

△ H_{f}^{\circ} H_{2} S = - 21.1

k J m o l^{- 1}

H_{2 (g)} \to 2 H_{(g)}

Δ H = 438 k J m o l^{- 1}

Δ H_{s u b l i m a t i o n}

280 k J m o l^{- 1}

△ H_{f}^{\circ} H S

C_{2} H_{2} (g) \to C_{2} H_{4} (g) : Δ H^{\circ} = - 175 {kJ mol}^{- 1}

Δ H_{f (C_{2} H_{4}, g)}^{\circ} = 50 kJ mo l^{- 1}; Δ H_{f (H_{2} O, l)}^{\circ} = - 280 {kJ mol}^{- 1}; Δ H_{f (C O_{2} g)}^{\circ} = - 390 {kJ mol}^{- 1}

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121 k J m o l^{- 1}

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242.8 k J m o l^{- 1}

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2422 k J m o l^{- 1}

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Q. Consider the following reactions:

(i) H^{+} (a q) + O H^{-} (a q) \to H_{2} O (l), Δ H = - X_{1} {kJ mol}^{- 1}

(i i) H_{2} (g) + \frac{1}{2} O_{2} (a q) \to H_{2} O (l), Δ H = - X_{2} {kJ mol}^{- 1}

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Heat of Reaction

Standard XIII Chemistry

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