QBManager

Thermodynamics

273018 If the mass defect of $_{5} B^{11}$ is $0.081 u$ , its average binding energy (in $MeV$ ) is :

1 8.60

2 6.85

3 5.60

4 5.86

Explanation:

Given: $Δ m = 0.081 u$ .
Number of nucleons $= 11$
$∴$ Binding energy $= Δ m \times 931 MeV$
$= 0.081 \times 931$
$= 75.411 MeV$
and average binding energy
$\begin{aligned} = \frac{Binding energy}{Number of nucleons} \\ = \frac{75.411}{11} \\ = 6.85 MeV . \end{aligned}$

A.P.EAMCET/2008

Thermodynamics

273019 Given, the bond energies of $N \equiv N, H - H$ and $N - H$ bonds as 945,436 and $391 kJ / mol$ respectively, the enthalpy of the reaction

1

- 93 kJ

2

102 kJ

3

90 kJ

4

105 kJ

Explanation:

Given, the reaction is
$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}$
Bond energy of $N \equiv N = 945 kJ / mole$ .
Bond energy of $3 (H - H) = 3 \times 436 = 1308 kJ / mole$
Bond energy of $6 (N - H) = 6 \times 391 = 2346 kJ / mole$
$∴$ Enthalpy of reaction $=$ reactant - product
$\begin{aligned} = 945 + 1308 - 2346 \\ = 2253 - 2346 \\ = - 93 {kJmol}^{- 1} \end{aligned}$

(A.P.EAMCET-1992)

Thermodynamics

273020 Average $C - H$ bond energy is $416 kJ {mol}^{- 1}$ . Which of the following is correct?

1

{CH}_{4} (g) + 416 kJ ⟶ C (g) + 4 H (g)

2

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 416 kJ

3

{CH}_{4} (g) + 1664 kJ ⟶ C (g) + 4 H (g)

4

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 1664 kJ

Explanation:

${CH}_{4}$ consists of $C - H$ bonds.
To break ${CH}_{4}$ into $C$ and $4 H$ , the energy required $=$
$4 \times 416 = 1664 kJ$
$∴$ The requirement satisfied by the following.
${CH}_{4} (g) + 1664 kJ \to C (g) + 4 H (g)$

(A.P.EAMCET-2004)

Thermodynamics

273022 Find the $Δ H^{0}$ for the reaction $H_{2} O (g)$ $+ {Br}_{2} (g) \to HBr (g) + HOBr (g)$ using:
Bond energy of $O - H$ bond in Water $= 463 kJ$
Bond energy of $Br - Br$ bond in ${Br}_{2} = 192 kJ$
Bond energy of $O - Br$ bond in $HOBr = 234 kJ$
Bond energy of $H - Br$ bond in $HBr = 364 kJ$

1

655 kJ . {mol}^{- 1}

2

598 kJ . {mol}^{- 1}

3

57 kJ . {mol}^{- 1}

4

192 kJ \cdot {mol}^{- 1}

Explanation:

The heat of reaction can be calculated based on the standard heat of formation of all reactants involved.
$\begin{aligned} Δ H^{\circ} & = [(234 + 364) - (463 + 192)] \\ = 598 - 655 \\ = - 57 kJ {mol}^{- 1} \end{aligned}$

Shift-II

Thermodynamics

273018 If the mass defect of $_{5} B^{11}$ is $0.081 u$ , its average binding energy (in $MeV$ ) is :

1 8.60

2 6.85

3 5.60

4 5.86

Explanation:

Given: $Δ m = 0.081 u$ .
Number of nucleons $= 11$
$∴$ Binding energy $= Δ m \times 931 MeV$
$= 0.081 \times 931$
$= 75.411 MeV$
and average binding energy
$\begin{aligned} = \frac{Binding energy}{Number of nucleons} \\ = \frac{75.411}{11} \\ = 6.85 MeV . \end{aligned}$

A.P.EAMCET/2008

Thermodynamics

273019 Given, the bond energies of $N \equiv N, H - H$ and $N - H$ bonds as 945,436 and $391 kJ / mol$ respectively, the enthalpy of the reaction

1

- 93 kJ

2

102 kJ

3

90 kJ

4

105 kJ

Explanation:

Given, the reaction is
$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}$
Bond energy of $N \equiv N = 945 kJ / mole$ .
Bond energy of $3 (H - H) = 3 \times 436 = 1308 kJ / mole$
Bond energy of $6 (N - H) = 6 \times 391 = 2346 kJ / mole$
$∴$ Enthalpy of reaction $=$ reactant - product
$\begin{aligned} = 945 + 1308 - 2346 \\ = 2253 - 2346 \\ = - 93 {kJmol}^{- 1} \end{aligned}$

(A.P.EAMCET-1992)

Thermodynamics

273020 Average $C - H$ bond energy is $416 kJ {mol}^{- 1}$ . Which of the following is correct?

1

{CH}_{4} (g) + 416 kJ ⟶ C (g) + 4 H (g)

2

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 416 kJ

3

{CH}_{4} (g) + 1664 kJ ⟶ C (g) + 4 H (g)

4

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 1664 kJ

Explanation:

${CH}_{4}$ consists of $C - H$ bonds.
To break ${CH}_{4}$ into $C$ and $4 H$ , the energy required $=$
$4 \times 416 = 1664 kJ$
$∴$ The requirement satisfied by the following.
${CH}_{4} (g) + 1664 kJ \to C (g) + 4 H (g)$

(A.P.EAMCET-2004)

Thermodynamics

273022 Find the $Δ H^{0}$ for the reaction $H_{2} O (g)$ $+ {Br}_{2} (g) \to HBr (g) + HOBr (g)$ using:
Bond energy of $O - H$ bond in Water $= 463 kJ$
Bond energy of $Br - Br$ bond in ${Br}_{2} = 192 kJ$
Bond energy of $O - Br$ bond in $HOBr = 234 kJ$
Bond energy of $H - Br$ bond in $HBr = 364 kJ$

1

655 kJ . {mol}^{- 1}

2

598 kJ . {mol}^{- 1}

3

57 kJ . {mol}^{- 1}

4

192 kJ \cdot {mol}^{- 1}

Explanation:

The heat of reaction can be calculated based on the standard heat of formation of all reactants involved.
$\begin{aligned} Δ H^{\circ} & = [(234 + 364) - (463 + 192)] \\ = 598 - 655 \\ = - 57 kJ {mol}^{- 1} \end{aligned}$

Shift-II

Thermodynamics

273018 If the mass defect of $_{5} B^{11}$ is $0.081 u$ , its average binding energy (in $MeV$ ) is :

1 8.60

2 6.85

3 5.60

4 5.86

Explanation:

Given: $Δ m = 0.081 u$ .
Number of nucleons $= 11$
$∴$ Binding energy $= Δ m \times 931 MeV$
$= 0.081 \times 931$
$= 75.411 MeV$
and average binding energy
$\begin{aligned} = \frac{Binding energy}{Number of nucleons} \\ = \frac{75.411}{11} \\ = 6.85 MeV . \end{aligned}$

A.P.EAMCET/2008

Thermodynamics

273019 Given, the bond energies of $N \equiv N, H - H$ and $N - H$ bonds as 945,436 and $391 kJ / mol$ respectively, the enthalpy of the reaction

1

- 93 kJ

2

102 kJ

3

90 kJ

4

105 kJ

Explanation:

Given, the reaction is
$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}$
Bond energy of $N \equiv N = 945 kJ / mole$ .
Bond energy of $3 (H - H) = 3 \times 436 = 1308 kJ / mole$
Bond energy of $6 (N - H) = 6 \times 391 = 2346 kJ / mole$
$∴$ Enthalpy of reaction $=$ reactant - product
$\begin{aligned} = 945 + 1308 - 2346 \\ = 2253 - 2346 \\ = - 93 {kJmol}^{- 1} \end{aligned}$

(A.P.EAMCET-1992)

Thermodynamics

273020 Average $C - H$ bond energy is $416 kJ {mol}^{- 1}$ . Which of the following is correct?

1

{CH}_{4} (g) + 416 kJ ⟶ C (g) + 4 H (g)

2

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 416 kJ

3

{CH}_{4} (g) + 1664 kJ ⟶ C (g) + 4 H (g)

4

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 1664 kJ

Explanation:

${CH}_{4}$ consists of $C - H$ bonds.
To break ${CH}_{4}$ into $C$ and $4 H$ , the energy required $=$
$4 \times 416 = 1664 kJ$
$∴$ The requirement satisfied by the following.
${CH}_{4} (g) + 1664 kJ \to C (g) + 4 H (g)$

(A.P.EAMCET-2004)

Thermodynamics

273022 Find the $Δ H^{0}$ for the reaction $H_{2} O (g)$ $+ {Br}_{2} (g) \to HBr (g) + HOBr (g)$ using:
Bond energy of $O - H$ bond in Water $= 463 kJ$
Bond energy of $Br - Br$ bond in ${Br}_{2} = 192 kJ$
Bond energy of $O - Br$ bond in $HOBr = 234 kJ$
Bond energy of $H - Br$ bond in $HBr = 364 kJ$

1

655 kJ . {mol}^{- 1}

2

598 kJ . {mol}^{- 1}

3

57 kJ . {mol}^{- 1}

4

192 kJ \cdot {mol}^{- 1}

Explanation:

The heat of reaction can be calculated based on the standard heat of formation of all reactants involved.
$\begin{aligned} Δ H^{\circ} & = [(234 + 364) - (463 + 192)] \\ = 598 - 655 \\ = - 57 kJ {mol}^{- 1} \end{aligned}$

Shift-II

Thermodynamics

273018 If the mass defect of $_{5} B^{11}$ is $0.081 u$ , its average binding energy (in $MeV$ ) is :

1 8.60

2 6.85

3 5.60

4 5.86

Explanation:

Given: $Δ m = 0.081 u$ .
Number of nucleons $= 11$
$∴$ Binding energy $= Δ m \times 931 MeV$
$= 0.081 \times 931$
$= 75.411 MeV$
and average binding energy
$\begin{aligned} = \frac{Binding energy}{Number of nucleons} \\ = \frac{75.411}{11} \\ = 6.85 MeV . \end{aligned}$

A.P.EAMCET/2008

Thermodynamics

273019 Given, the bond energies of $N \equiv N, H - H$ and $N - H$ bonds as 945,436 and $391 kJ / mol$ respectively, the enthalpy of the reaction

1

- 93 kJ

2

102 kJ

3

90 kJ

4

105 kJ

Explanation:

Given, the reaction is
$N_{2} + 3 H_{2} ⟶ 2 {NH}_{3}$
Bond energy of $N \equiv N = 945 kJ / mole$ .
Bond energy of $3 (H - H) = 3 \times 436 = 1308 kJ / mole$
Bond energy of $6 (N - H) = 6 \times 391 = 2346 kJ / mole$
$∴$ Enthalpy of reaction $=$ reactant - product
$\begin{aligned} = 945 + 1308 - 2346 \\ = 2253 - 2346 \\ = - 93 {kJmol}^{- 1} \end{aligned}$

(A.P.EAMCET-1992)

Thermodynamics

273020 Average $C - H$ bond energy is $416 kJ {mol}^{- 1}$ . Which of the following is correct?

1

{CH}_{4} (g) + 416 kJ ⟶ C (g) + 4 H (g)

2

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 416 kJ

3

{CH}_{4} (g) + 1664 kJ ⟶ C (g) + 4 H (g)

4

{CH}_{4} (g) ⟶ C (g) + 4 H (g) + 1664 kJ

Explanation:

${CH}_{4}$ consists of $C - H$ bonds.
To break ${CH}_{4}$ into $C$ and $4 H$ , the energy required $=$
$4 \times 416 = 1664 kJ$
$∴$ The requirement satisfied by the following.
${CH}_{4} (g) + 1664 kJ \to C (g) + 4 H (g)$

(A.P.EAMCET-2004)

Thermodynamics

273022 Find the $Δ H^{0}$ for the reaction $H_{2} O (g)$ $+ {Br}_{2} (g) \to HBr (g) + HOBr (g)$ using:
Bond energy of $O - H$ bond in Water $= 463 kJ$
Bond energy of $Br - Br$ bond in ${Br}_{2} = 192 kJ$
Bond energy of $O - Br$ bond in $HOBr = 234 kJ$
Bond energy of $H - Br$ bond in $HBr = 364 kJ$

1

655 kJ . {mol}^{- 1}

2

598 kJ . {mol}^{- 1}

3

57 kJ . {mol}^{- 1}

4

192 kJ \cdot {mol}^{- 1}

Explanation:

The heat of reaction can be calculated based on the standard heat of formation of all reactants involved.
$\begin{aligned} Δ H^{\circ} & = [(234 + 364) - (463 + 192)] \\ = 598 - 655 \\ = - 57 kJ {mol}^{- 1} \end{aligned}$

Shift-II

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