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CHXI06:THERMODYNAMICS

369530 Energy required to dissociate $4 g$ of gaseous hydrogen into free gaseous atoms is $208 kcal$ at $25^{\circ} C$ . The bond energy of $H - H$ bond will be

1

104 kcal

2

10.4 kcal

3

1040 kcal

4

- 104 kcal

Explanation:

$4 g H_{2} = 2$ moles. Bond energy for 1 mole of $H_{2} = 208 / 2 = 104 k c a l$

CHXI06:THERMODYNAMICS

369531 What will be the enthalpy change for the following reaction?
$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$
Given that Bond energy of $H_{2}, {Br}_{2}$ and $HBr$ is $435 kJ {mol}^{- 1}, 192 kJ {mol}^{- 1}$ and $368 kJ {mol}^{- 1}$ respectively.

1

+ 259 k J m o l^{- 1}

2

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

3

+ 109 k J m o l^{- 1}

4

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

Explanation:

$Δ_{r} H^{\circ} = \sum B . E_{r e a} - \sum B . E_{p r o}$
$Δ_{r} H^{\circ} =$ Bond energy of $H_{2} +$ Bond energy of ${Br}_{2} - 2 \times$ Bond energy of $HBr$
$= 435 + 192 - (2 \times 368) kJ {mol}^{- 1}$ $\Rightarrow Δ_{r} H^{\circ} = - 109 kJ {mol}^{- 1}$ .

CHXI06:THERMODYNAMICS

369532 The enthalpy changes at $298 K$ in successive breaking of $O - H$ bonds in $HOH$ are
$H_{2} O_{(g)} \to H_{(g)} + {OH}_{(g)}; Δ H = 498 kJ {mol}^{- 1}$
${OH}_{(g)} \to H_{(g)} + O_{(g)}; Δ H = 428 kJ {mol}^{- 1}$
The bond enthalpy of the $O - H$ bond is

1

498 kJ {mol}^{- 1}

2

463 kJ {mol}^{- 1}

3

428 kJ {mol}^{- 1}

4

70 kJ {mol}^{- 1}

Explanation:

Average value $= \frac{(498 + 428)}{2} = 463 kJ / mol$

CHXI06:THERMODYNAMICS

369533 The standard heat of formation values of ${SF}_{6} (g), S_{(s)}$ , and $F (g)$ are $- 1100, 275$ , and 80 $kJ / mol$ , respectively. Then the average $S - F$ bond energy in ${SF}_{6}$ is

1

301 kJ / mol

2

320 kJ / mol

3

309 kJ / mol

4

280 kJ / mol

Explanation:

$S_{(g)} + 6 F_{(g)} \to {SF}_{6 (g)}; Δ H_{f} = - 1100 kJ / mol$
$S_{(s)} ⟶ S_{(g)}; Δ H = + 275 kJ / mol$
$1 / 2$ $F_{2 (g)} \to F_{(g)}; Δ H = 80 k J / m o l$
$∴$ Heat of formation $=$ Bond energy of reactant
- Bond energy of product
$- 1100 = [275 + 6 \times 80] - [6 \times S - F]$
Thus, bond energy of $S - F = 309 kJ / mol$ .

CHXI06:THERMODYNAMICS

369530 Energy required to dissociate $4 g$ of gaseous hydrogen into free gaseous atoms is $208 kcal$ at $25^{\circ} C$ . The bond energy of $H - H$ bond will be

1

104 kcal

2

10.4 kcal

3

1040 kcal

4

- 104 kcal

Explanation:

$4 g H_{2} = 2$ moles. Bond energy for 1 mole of $H_{2} = 208 / 2 = 104 k c a l$

CHXI06:THERMODYNAMICS

369531 What will be the enthalpy change for the following reaction?
$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$
Given that Bond energy of $H_{2}, {Br}_{2}$ and $HBr$ is $435 kJ {mol}^{- 1}, 192 kJ {mol}^{- 1}$ and $368 kJ {mol}^{- 1}$ respectively.

1

+ 259 k J m o l^{- 1}

2

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

3

+ 109 k J m o l^{- 1}

4

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

Explanation:

$Δ_{r} H^{\circ} = \sum B . E_{r e a} - \sum B . E_{p r o}$
$Δ_{r} H^{\circ} =$ Bond energy of $H_{2} +$ Bond energy of ${Br}_{2} - 2 \times$ Bond energy of $HBr$
$= 435 + 192 - (2 \times 368) kJ {mol}^{- 1}$ $\Rightarrow Δ_{r} H^{\circ} = - 109 kJ {mol}^{- 1}$ .

CHXI06:THERMODYNAMICS

369532 The enthalpy changes at $298 K$ in successive breaking of $O - H$ bonds in $HOH$ are
$H_{2} O_{(g)} \to H_{(g)} + {OH}_{(g)}; Δ H = 498 kJ {mol}^{- 1}$
${OH}_{(g)} \to H_{(g)} + O_{(g)}; Δ H = 428 kJ {mol}^{- 1}$
The bond enthalpy of the $O - H$ bond is

1

498 kJ {mol}^{- 1}

2

463 kJ {mol}^{- 1}

3

428 kJ {mol}^{- 1}

4

70 kJ {mol}^{- 1}

Explanation:

Average value $= \frac{(498 + 428)}{2} = 463 kJ / mol$

CHXI06:THERMODYNAMICS

369533 The standard heat of formation values of ${SF}_{6} (g), S_{(s)}$ , and $F (g)$ are $- 1100, 275$ , and 80 $kJ / mol$ , respectively. Then the average $S - F$ bond energy in ${SF}_{6}$ is

1

301 kJ / mol

2

320 kJ / mol

3

309 kJ / mol

4

280 kJ / mol

Explanation:

$S_{(g)} + 6 F_{(g)} \to {SF}_{6 (g)}; Δ H_{f} = - 1100 kJ / mol$
$S_{(s)} ⟶ S_{(g)}; Δ H = + 275 kJ / mol$
$1 / 2$ $F_{2 (g)} \to F_{(g)}; Δ H = 80 k J / m o l$
$∴$ Heat of formation $=$ Bond energy of reactant
- Bond energy of product
$- 1100 = [275 + 6 \times 80] - [6 \times S - F]$
Thus, bond energy of $S - F = 309 kJ / mol$ .

CHXI06:THERMODYNAMICS

369530 Energy required to dissociate $4 g$ of gaseous hydrogen into free gaseous atoms is $208 kcal$ at $25^{\circ} C$ . The bond energy of $H - H$ bond will be

1

104 kcal

2

10.4 kcal

3

1040 kcal

4

- 104 kcal

Explanation:

$4 g H_{2} = 2$ moles. Bond energy for 1 mole of $H_{2} = 208 / 2 = 104 k c a l$

CHXI06:THERMODYNAMICS

369531 What will be the enthalpy change for the following reaction?
$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$
Given that Bond energy of $H_{2}, {Br}_{2}$ and $HBr$ is $435 kJ {mol}^{- 1}, 192 kJ {mol}^{- 1}$ and $368 kJ {mol}^{- 1}$ respectively.

1

+ 259 k J m o l^{- 1}

2

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

3

+ 109 k J m o l^{- 1}

4

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

Explanation:

$Δ_{r} H^{\circ} = \sum B . E_{r e a} - \sum B . E_{p r o}$
$Δ_{r} H^{\circ} =$ Bond energy of $H_{2} +$ Bond energy of ${Br}_{2} - 2 \times$ Bond energy of $HBr$
$= 435 + 192 - (2 \times 368) kJ {mol}^{- 1}$ $\Rightarrow Δ_{r} H^{\circ} = - 109 kJ {mol}^{- 1}$ .

CHXI06:THERMODYNAMICS

369532 The enthalpy changes at $298 K$ in successive breaking of $O - H$ bonds in $HOH$ are
$H_{2} O_{(g)} \to H_{(g)} + {OH}_{(g)}; Δ H = 498 kJ {mol}^{- 1}$
${OH}_{(g)} \to H_{(g)} + O_{(g)}; Δ H = 428 kJ {mol}^{- 1}$
The bond enthalpy of the $O - H$ bond is

1

498 kJ {mol}^{- 1}

2

463 kJ {mol}^{- 1}

3

428 kJ {mol}^{- 1}

4

70 kJ {mol}^{- 1}

Explanation:

Average value $= \frac{(498 + 428)}{2} = 463 kJ / mol$

CHXI06:THERMODYNAMICS

369533 The standard heat of formation values of ${SF}_{6} (g), S_{(s)}$ , and $F (g)$ are $- 1100, 275$ , and 80 $kJ / mol$ , respectively. Then the average $S - F$ bond energy in ${SF}_{6}$ is

1

301 kJ / mol

2

320 kJ / mol

3

309 kJ / mol

4

280 kJ / mol

Explanation:

$S_{(g)} + 6 F_{(g)} \to {SF}_{6 (g)}; Δ H_{f} = - 1100 kJ / mol$
$S_{(s)} ⟶ S_{(g)}; Δ H = + 275 kJ / mol$
$1 / 2$ $F_{2 (g)} \to F_{(g)}; Δ H = 80 k J / m o l$
$∴$ Heat of formation $=$ Bond energy of reactant
- Bond energy of product
$- 1100 = [275 + 6 \times 80] - [6 \times S - F]$
Thus, bond energy of $S - F = 309 kJ / mol$ .

CHXI06:THERMODYNAMICS

369530 Energy required to dissociate $4 g$ of gaseous hydrogen into free gaseous atoms is $208 kcal$ at $25^{\circ} C$ . The bond energy of $H - H$ bond will be

1

104 kcal

2

10.4 kcal

3

1040 kcal

4

- 104 kcal

Explanation:

$4 g H_{2} = 2$ moles. Bond energy for 1 mole of $H_{2} = 208 / 2 = 104 k c a l$

CHXI06:THERMODYNAMICS

369531 What will be the enthalpy change for the following reaction?
$H_{2} (g) + B r_{2} (g) \to 2 H B r (g)$
Given that Bond energy of $H_{2}, {Br}_{2}$ and $HBr$ is $435 kJ {mol}^{- 1}, 192 kJ {mol}^{- 1}$ and $368 kJ {mol}^{- 1}$ respectively.

1

+ 259 k J m o l^{- 1}

2

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

3

+ 109 k J m o l^{- 1}

4

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

Explanation:

$Δ_{r} H^{\circ} = \sum B . E_{r e a} - \sum B . E_{p r o}$
$Δ_{r} H^{\circ} =$ Bond energy of $H_{2} +$ Bond energy of ${Br}_{2} - 2 \times$ Bond energy of $HBr$
$= 435 + 192 - (2 \times 368) kJ {mol}^{- 1}$ $\Rightarrow Δ_{r} H^{\circ} = - 109 kJ {mol}^{- 1}$ .

CHXI06:THERMODYNAMICS

369532 The enthalpy changes at $298 K$ in successive breaking of $O - H$ bonds in $HOH$ are
$H_{2} O_{(g)} \to H_{(g)} + {OH}_{(g)}; Δ H = 498 kJ {mol}^{- 1}$
${OH}_{(g)} \to H_{(g)} + O_{(g)}; Δ H = 428 kJ {mol}^{- 1}$
The bond enthalpy of the $O - H$ bond is

1

498 kJ {mol}^{- 1}

2

463 kJ {mol}^{- 1}

3

428 kJ {mol}^{- 1}

4

70 kJ {mol}^{- 1}

Explanation:

Average value $= \frac{(498 + 428)}{2} = 463 kJ / mol$

CHXI06:THERMODYNAMICS

369533 The standard heat of formation values of ${SF}_{6} (g), S_{(s)}$ , and $F (g)$ are $- 1100, 275$ , and 80 $kJ / mol$ , respectively. Then the average $S - F$ bond energy in ${SF}_{6}$ is

1

301 kJ / mol

2

320 kJ / mol

3

309 kJ / mol

4

280 kJ / mol

Explanation:

$S_{(g)} + 6 F_{(g)} \to {SF}_{6 (g)}; Δ H_{f} = - 1100 kJ / mol$
$S_{(s)} ⟶ S_{(g)}; Δ H = + 275 kJ / mol$
$1 / 2$ $F_{2 (g)} \to F_{(g)}; Δ H = 80 k J / m o l$
$∴$ Heat of formation $=$ Bond energy of reactant
- Bond energy of product
$- 1100 = [275 + 6 \times 80] - [6 \times S - F]$
Thus, bond energy of $S - F = 309 kJ / mol$ .

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