If 2Als - 1 1 - 2 O2g- Al2O3s- - H - -166-78 kJ H2g - 1 - 2 O2g- H2Ol- - H - -285-9 kJThen - H for the reaction- Al2O3s- 2Als - 1 1 - 2 O2g is-

The correct option is D Δ H = 166.78 kJ mol^ 1 As given, 2Al(s) + 1 1 / 2 O_2(g)⟶ Al_2O_3(s), Δ H = 166.78 kJ and by this we can see that,for revers ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

If 2Als + 1...

Question

If $2 A l (s) + 1 \frac{1}{2} O_{2} (g) ⟶ A l_{2} O_{3} (s), Δ H = - 166.78 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l), Δ H = - 285.9 k J$
Then $Δ H$ for the reaction,
$A l_{2} O_{3} (s) ⟶ 2 A l (s) + 1 \frac{1}{2} O_{2} (g)$ is:

Δ H = 174.20 k J m o l^{- 1}

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Δ H = 124.55 k J m o l^{- 1}

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Δ H = 166.78 k J m o l^{- 1}

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Δ H = 156.33 k J m o l^{- 1}

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Solution

The correct option is D $Δ H = 166.78 k J m o l^{- 1}$
As given,
$2 A l (s) + 1 \frac{1}{2} O_{2} (g) ⟶ A l_{2} O_{3} (s), Δ H = - 166.78 k J$
and by this we can see that,
for reverse reaction
$A l_{2} O_{3} (s) ⟶ 2 A l (s) + 1 \frac{1}{2} O_{2} (g) Δ H = 166.78 k J$

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Similar questions

Q. Calculate

Δ H

k J

for the following reaction:

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

Given that,

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

;

Δ H = + 131 k J

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

;

Δ H = - 282 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)

;

Δ H = - 242 k J

Q. Given that:

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g), Δ H = x k J m o l^{- 1}

C (g r a p h i t e) + \frac{1}{2} O_{2} (g) \to C O (g), Δ H = y k J m o l^{- 1}

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ H = z k J m o l^{- 1}

Based on the given thermochemical equations, find out which one of the following algebric relationships is correct?

2 B (s) + \frac{3}{2} O_2 (g) \to B_{2} O_{3} (s .)

Δ H = - 1273 k j

H_2 (g) + \frac{1}{2} O_2 (g) \to H_2 O (l)

Δ H = - 286 k j

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

Δ H = 44 k j

2 B (s) + 3 H_{2} (g) \to B_{2} H_{6} (g)

Δ H = 365 k j

B_{2} H_{6} (g) + 3 O_{2} (g) B_{2} O_{3} (s) + 3 H_{2} O (g)

Δ H = ?

Q. Consider the following reactions :

H^{+} (a q) + O H^{-} (a q) = H_{2} O (l); Δ H = - X_{1} k J . m o l^{- 1}

II)

H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (l); Δ H = - X_{2} k J . m o l^{- 1}

III)

C O_{2} (g) + H_{2} (g) = C O (g) + H_{2} O (l); Δ H = + X_{3} k J . m o l^{- 1}

IV)

C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) = 2 C O (g) + H_{2} O (l); Δ H = + X_{4} k J . m o l^{- 1}

Enthalpy of formation of

H_{2} O (l)

is:

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = x

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = y

Heat of vaporisation of water is:

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If 2Al(s)+112O2(g)⟶Al2O3(s),ΔH=−166.78kJH2(g)+12O2(g)⟶H2O(l),ΔH=−285.9kJThen ΔH for the reaction,Al2O3(s)⟶2Al(s)+112O2(g) is:

The correct option is D ΔH=166.78kJmol−1As given,2Al(s)+112O2(g)⟶Al2O3(s),ΔH=−166.78kJand by this we can see that,for reverse reactionAl2O3(s)⟶2Al(s)+112O2(g)ΔH=166.78kJ

If $2 A l (s) + 1 \frac{1}{2} O_{2} (g) ⟶ A l_{2} O_{3} (s), Δ H = - 166.78 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l), Δ H = - 285.9 k J$
Then $Δ H$ for the reaction,
$A l_{2} O_{3} (s) ⟶ 2 A l (s) + 1 \frac{1}{2} O_{2} (g)$ is:

The correct option is D $Δ H = 166.78 k J m o l^{- 1}$
As given,
$2 A l (s) + 1 \frac{1}{2} O_{2} (g) ⟶ A l_{2} O_{3} (s), Δ H = - 166.78 k J$
and by this we can see that,
for reverse reaction
$A l_{2} O_{3} (s) ⟶ 2 A l (s) + 1 \frac{1}{2} O_{2} (g) Δ H = 166.78 k J$