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Standard XII

Chemistry

Enthalpy

For the react...

Question

For the reaction,

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$ ; $Δ H = - 571 k J$ .

Bond energies: $(H - H) = 435 k J$ , $(O = O) = 498 k J$ .

Calculate the average bond energy of $(O - H)$ using the above data.

484 kJ

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-484 kJ

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271 kJ

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-271 kJ

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Solution

The correct option is A
484 kJ

$2 (H - H) + O = O \to 2 (H - O - H)$

$Δ H = \sum (B E)_{reactants} - \sum (B E)_{products}$

$- 571 = [2 \times 435 + 498] - 4 \times (B E)_{O - H}$

$(B E)_{O - H} = \frac{(2 \times 435) + 498 + 571}{4} \approx 484 k J$

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Similar questions

Q. Use the bond energies to estimate

Δ H

for this reaction:

H_{2} (g) + O_{2} (g) ⟶ H_{2} O_{2} (g)

Bond

Bond energy

H - H

436 k J

{m o l}^{- 1}

O - O

142 k J

{m o l}^{- 1}

O = O

499 k J

{m o l}^{- 1}

H - O

460 k J

{m o l}^{- 1}

Q. Find the

Δ H

of the following reaction.

O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)

. Average bond energies of

O - F

O - H

O = O

and

H - F

are

44

111

118

and

135

kcal mol

^{- 1}

, respectively.

Q. Calculate the enthalpy changes

(δ H)

of the following reaction.

2 C_{2} H_{2 (g)} + 5 O_{2 (g)} \to 4 C O_{2 (g)} + 2 H_{2} O_{(g)}

given average bond enthalpies of various bonds, i.e

C - H, C \equiv C, O = O, C = O, O - H

414, 814, 499, 724

and

640 k J m o l^{- 1}

respectively.

Q. The average O-H bond energy with the help of following data is:

(1) 2 H (g) ⟶ H_{2} (g); Δ H = - 435.0 {kJ mol}^{- 1}

(2) O_{2} (g) ⟶ 2 O (g); Δ H = + 489.6 {kJ mol}^{- 1}

(3) \frac{1}{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ O H; Δ H = - 42.09 {kJ mol}^{- 1}

Q. For the reaction,

O (g) + O_{2} (g) \to O_{3} (g); Δ H^{o} = - 107.2 k J

. Hence, the average bond energy in

O_{3}

is:

(Assume

O = O

bond energy

498.8 k J m o l^{- 1}

)

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For the reaction, 2H2(g)+O2(g)→2H2O(g); ΔH=−571 kJ. Bond energies: (H−H)=435 kJ, (O=O)=498 kJ. Calculate the average bond energy of (O−H) using the above data.

The correct option is A 484 kJ 2(H−H)+O=O→2(H−O−H) ΔH=∑(BE)reactants−∑(BE)products −571=[2×435+498]−4×(BE)O−H (BE)O−H=(2×435)+498+5714≈484 kJ

For the reaction,

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$ ; $Δ H = - 571 k J$ .

Bond energies: $(H - H) = 435 k J$ , $(O = O) = 498 k J$ .

Calculate the average bond energy of $(O - H)$ using the above data.

The correct option is A
484 kJ

$2 (H - H) + O = O \to 2 (H - O - H)$

$Δ H = \sum (B E)_{reactants} - \sum (B E)_{products}$

$- 571 = [2 \times 435 + 498] - 4 \times (B E)_{O - H}$

$(B E)_{O - H} = \frac{(2 \times 435) + 498 + 571}{4} \approx 484 k J$