QBManager

Thermodynamics

273048 If the bond energies of $H - H, Br - Br$ , and $H -$ $B r$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, the $Δ H^{0}$ for the reaction $H_{2} (g) + {Br}_{2} (g) \to$ $2 HBr (g)$ is

1

- 261 kJ

2

+ 103 kJ

3

+ 261 kJ

4

- 103 kJ

Explanation:

Given, Bond energy -
$\begin{aligned} {BE}_{H - H} = 433 kJ / mole = Δ H \\ {BE}_{Br - Br} = 192 kJ / Mole = Δ H \\ {BE}_{H - Br} = 364 kJ / mole = Δ H \\ H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g) \end{aligned}$
$Δ H^{\circ}$ for the reaction
$\begin{aligned} Δ H^{\circ} = Σ B \cdot E_{R} - Σ B \cdot E_{P} \\ = & ({BE}_{(H - H)} + {BE}_{(Br - Br)}) - 2 {BE}_{(H - Br)} \\ = & (433 + 192) - 2 (364) \\ = & 625 - 728 \end{aligned}$
$Δ H^{\circ} = - 103 kJ$

[Karnataka-CET-2016]

Thermodynamics

273049 Given that bond energies of $H - H$ and $Cl - Cl$ are $430 kJ {mol}^{- 1}$ and $240 kJ {mol}^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- 90 kJ {mol}^{- 1}$ , bond enthalpy of $HCl$ is

1

380 kJ {mol}^{- 1}

2

425 kJ {mol}^{- 1}

3

245 kJ {mol}^{- 1}

4

290 kJ {mol}^{- 1}

Explanation:

$H_{2} + {Cl}_{2} \to 2 HCl$
${BE}_{(H - H)} = 430 kJ / mol$
${BE}_{(Cl - Cl)} = 240 kJ / mol$
$Δ H_{(fHCl)} = - 70 kJ / mol$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$
$Δ H_{f} = - 90 kJ / mol$
$\begin{aligned} Δ H_{f} = \frac{1}{2} {BE}_{(H - H)} + \frac{1}{2} {BE}_{(Cl - Cl)} - {BE}_{(HCl)} \\ - 90 = \frac{1}{2} \times 430 + \frac{1}{2} \times 240 - {BE}_{(HCl)} \\ {BE}_{(HCl)} = 215 + 120 + 90 \\ = 425 kJ / mol \end{aligned}$

NEET-2007

Thermodynamics

273050 For the reaction
$A (g) + 2 B (g) ⟶ 2 C (g) - 3 D (g)$
the change of enthalpy at $27^{\circ} C$ is 19 kcal. The value of $Δ E$ is: $(R = 2.02 cal K^{- 1} {mol}^{- 1})$

1

21.2 kcal

2

17.8 kcal

3

18.4 kcal

4

20.6 kcal

Explanation:

We know that, $Δ H = Δ E + Δ n_{g} RT$
$\begin{aligned} Δ n_{g} = 5 - 3 = 2 \\ Δ H = 19 kcal \\ = 19 \times 10^{3} cal . \\ Δ E = Δ H - Δ n_{g} RT \\ = 19 \times 10^{3} - (2 \times 2 \times 300) \\ = 19000 - 1200 = 17800 cal \\ ⊔ 17.8 kcal \end{aligned}$

UPTU/UPSEE-2006

Thermodynamics

273037 At $298 K$ temperature the activation energy for the reaction $x_{2} + y_{2} \to 2 x y + 20 k J$ is $15 kJ$ .
What will be the activation energy for the reaction $2 x y \to x_{2} + y_{2}$ ?

1

- 15 kJ

2

+ 35 kJ

3

- 5 kJ

4

- 35 kJ

Explanation:

$Δ H_{R} =$ activation energy of forward reactionactivation energy of backward reaction-
$\begin{array}{ll} ∵ & - 20 = 15 - x \\ So, & x = + 35 kJ \end{array}$

GUJCET-2017

Thermodynamics

273048 If the bond energies of $H - H, Br - Br$ , and $H -$ $B r$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, the $Δ H^{0}$ for the reaction $H_{2} (g) + {Br}_{2} (g) \to$ $2 HBr (g)$ is

1

- 261 kJ

2

+ 103 kJ

3

+ 261 kJ

4

- 103 kJ

Explanation:

Given, Bond energy -
$\begin{aligned} {BE}_{H - H} = 433 kJ / mole = Δ H \\ {BE}_{Br - Br} = 192 kJ / Mole = Δ H \\ {BE}_{H - Br} = 364 kJ / mole = Δ H \\ H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g) \end{aligned}$
$Δ H^{\circ}$ for the reaction
$\begin{aligned} Δ H^{\circ} = Σ B \cdot E_{R} - Σ B \cdot E_{P} \\ = & ({BE}_{(H - H)} + {BE}_{(Br - Br)}) - 2 {BE}_{(H - Br)} \\ = & (433 + 192) - 2 (364) \\ = & 625 - 728 \end{aligned}$
$Δ H^{\circ} = - 103 kJ$

[Karnataka-CET-2016]

Thermodynamics

273049 Given that bond energies of $H - H$ and $Cl - Cl$ are $430 kJ {mol}^{- 1}$ and $240 kJ {mol}^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- 90 kJ {mol}^{- 1}$ , bond enthalpy of $HCl$ is

1

380 kJ {mol}^{- 1}

2

425 kJ {mol}^{- 1}

3

245 kJ {mol}^{- 1}

4

290 kJ {mol}^{- 1}

Explanation:

$H_{2} + {Cl}_{2} \to 2 HCl$
${BE}_{(H - H)} = 430 kJ / mol$
${BE}_{(Cl - Cl)} = 240 kJ / mol$
$Δ H_{(fHCl)} = - 70 kJ / mol$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$
$Δ H_{f} = - 90 kJ / mol$
$\begin{aligned} Δ H_{f} = \frac{1}{2} {BE}_{(H - H)} + \frac{1}{2} {BE}_{(Cl - Cl)} - {BE}_{(HCl)} \\ - 90 = \frac{1}{2} \times 430 + \frac{1}{2} \times 240 - {BE}_{(HCl)} \\ {BE}_{(HCl)} = 215 + 120 + 90 \\ = 425 kJ / mol \end{aligned}$

NEET-2007

Thermodynamics

273050 For the reaction
$A (g) + 2 B (g) ⟶ 2 C (g) - 3 D (g)$
the change of enthalpy at $27^{\circ} C$ is 19 kcal. The value of $Δ E$ is: $(R = 2.02 cal K^{- 1} {mol}^{- 1})$

1

21.2 kcal

2

17.8 kcal

3

18.4 kcal

4

20.6 kcal

Explanation:

We know that, $Δ H = Δ E + Δ n_{g} RT$
$\begin{aligned} Δ n_{g} = 5 - 3 = 2 \\ Δ H = 19 kcal \\ = 19 \times 10^{3} cal . \\ Δ E = Δ H - Δ n_{g} RT \\ = 19 \times 10^{3} - (2 \times 2 \times 300) \\ = 19000 - 1200 = 17800 cal \\ ⊔ 17.8 kcal \end{aligned}$

UPTU/UPSEE-2006

Thermodynamics

273037 At $298 K$ temperature the activation energy for the reaction $x_{2} + y_{2} \to 2 x y + 20 k J$ is $15 kJ$ .
What will be the activation energy for the reaction $2 x y \to x_{2} + y_{2}$ ?

1

- 15 kJ

2

+ 35 kJ

3

- 5 kJ

4

- 35 kJ

Explanation:

$Δ H_{R} =$ activation energy of forward reactionactivation energy of backward reaction-
$\begin{array}{ll} ∵ & - 20 = 15 - x \\ So, & x = + 35 kJ \end{array}$

GUJCET-2017

Thermodynamics

273048 If the bond energies of $H - H, Br - Br$ , and $H -$ $B r$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, the $Δ H^{0}$ for the reaction $H_{2} (g) + {Br}_{2} (g) \to$ $2 HBr (g)$ is

1

- 261 kJ

2

+ 103 kJ

3

+ 261 kJ

4

- 103 kJ

Explanation:

Given, Bond energy -
$\begin{aligned} {BE}_{H - H} = 433 kJ / mole = Δ H \\ {BE}_{Br - Br} = 192 kJ / Mole = Δ H \\ {BE}_{H - Br} = 364 kJ / mole = Δ H \\ H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g) \end{aligned}$
$Δ H^{\circ}$ for the reaction
$\begin{aligned} Δ H^{\circ} = Σ B \cdot E_{R} - Σ B \cdot E_{P} \\ = & ({BE}_{(H - H)} + {BE}_{(Br - Br)}) - 2 {BE}_{(H - Br)} \\ = & (433 + 192) - 2 (364) \\ = & 625 - 728 \end{aligned}$
$Δ H^{\circ} = - 103 kJ$

[Karnataka-CET-2016]

Thermodynamics

273049 Given that bond energies of $H - H$ and $Cl - Cl$ are $430 kJ {mol}^{- 1}$ and $240 kJ {mol}^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- 90 kJ {mol}^{- 1}$ , bond enthalpy of $HCl$ is

1

380 kJ {mol}^{- 1}

2

425 kJ {mol}^{- 1}

3

245 kJ {mol}^{- 1}

4

290 kJ {mol}^{- 1}

Explanation:

$H_{2} + {Cl}_{2} \to 2 HCl$
${BE}_{(H - H)} = 430 kJ / mol$
${BE}_{(Cl - Cl)} = 240 kJ / mol$
$Δ H_{(fHCl)} = - 70 kJ / mol$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$
$Δ H_{f} = - 90 kJ / mol$
$\begin{aligned} Δ H_{f} = \frac{1}{2} {BE}_{(H - H)} + \frac{1}{2} {BE}_{(Cl - Cl)} - {BE}_{(HCl)} \\ - 90 = \frac{1}{2} \times 430 + \frac{1}{2} \times 240 - {BE}_{(HCl)} \\ {BE}_{(HCl)} = 215 + 120 + 90 \\ = 425 kJ / mol \end{aligned}$

NEET-2007

Thermodynamics

273050 For the reaction
$A (g) + 2 B (g) ⟶ 2 C (g) - 3 D (g)$
the change of enthalpy at $27^{\circ} C$ is 19 kcal. The value of $Δ E$ is: $(R = 2.02 cal K^{- 1} {mol}^{- 1})$

1

21.2 kcal

2

17.8 kcal

3

18.4 kcal

4

20.6 kcal

Explanation:

We know that, $Δ H = Δ E + Δ n_{g} RT$
$\begin{aligned} Δ n_{g} = 5 - 3 = 2 \\ Δ H = 19 kcal \\ = 19 \times 10^{3} cal . \\ Δ E = Δ H - Δ n_{g} RT \\ = 19 \times 10^{3} - (2 \times 2 \times 300) \\ = 19000 - 1200 = 17800 cal \\ ⊔ 17.8 kcal \end{aligned}$

UPTU/UPSEE-2006

Thermodynamics

273037 At $298 K$ temperature the activation energy for the reaction $x_{2} + y_{2} \to 2 x y + 20 k J$ is $15 kJ$ .
What will be the activation energy for the reaction $2 x y \to x_{2} + y_{2}$ ?

1

- 15 kJ

2

+ 35 kJ

3

- 5 kJ

4

- 35 kJ

Explanation:

$Δ H_{R} =$ activation energy of forward reactionactivation energy of backward reaction-
$\begin{array}{ll} ∵ & - 20 = 15 - x \\ So, & x = + 35 kJ \end{array}$

GUJCET-2017

Thermodynamics

273048 If the bond energies of $H - H, Br - Br$ , and $H -$ $B r$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, the $Δ H^{0}$ for the reaction $H_{2} (g) + {Br}_{2} (g) \to$ $2 HBr (g)$ is

1

- 261 kJ

2

+ 103 kJ

3

+ 261 kJ

4

- 103 kJ

Explanation:

Given, Bond energy -
$\begin{aligned} {BE}_{H - H} = 433 kJ / mole = Δ H \\ {BE}_{Br - Br} = 192 kJ / Mole = Δ H \\ {BE}_{H - Br} = 364 kJ / mole = Δ H \\ H_{2} (g) + {Br}_{2} (g) \to 2 HBr (g) \end{aligned}$
$Δ H^{\circ}$ for the reaction
$\begin{aligned} Δ H^{\circ} = Σ B \cdot E_{R} - Σ B \cdot E_{P} \\ = & ({BE}_{(H - H)} + {BE}_{(Br - Br)}) - 2 {BE}_{(H - Br)} \\ = & (433 + 192) - 2 (364) \\ = & 625 - 728 \end{aligned}$
$Δ H^{\circ} = - 103 kJ$

[Karnataka-CET-2016]

Thermodynamics

273049 Given that bond energies of $H - H$ and $Cl - Cl$ are $430 kJ {mol}^{- 1}$ and $240 kJ {mol}^{- 1}$ respectively and $Δ H_{f}$ for $HCl$ is $- 90 kJ {mol}^{- 1}$ , bond enthalpy of $HCl$ is

1

380 kJ {mol}^{- 1}

2

425 kJ {mol}^{- 1}

3

245 kJ {mol}^{- 1}

4

290 kJ {mol}^{- 1}

Explanation:

$H_{2} + {Cl}_{2} \to 2 HCl$
${BE}_{(H - H)} = 430 kJ / mol$
${BE}_{(Cl - Cl)} = 240 kJ / mol$
$Δ H_{(fHCl)} = - 70 kJ / mol$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$
$Δ H_{f} = - 90 kJ / mol$
$\begin{aligned} Δ H_{f} = \frac{1}{2} {BE}_{(H - H)} + \frac{1}{2} {BE}_{(Cl - Cl)} - {BE}_{(HCl)} \\ - 90 = \frac{1}{2} \times 430 + \frac{1}{2} \times 240 - {BE}_{(HCl)} \\ {BE}_{(HCl)} = 215 + 120 + 90 \\ = 425 kJ / mol \end{aligned}$

NEET-2007

Thermodynamics

273050 For the reaction
$A (g) + 2 B (g) ⟶ 2 C (g) - 3 D (g)$
the change of enthalpy at $27^{\circ} C$ is 19 kcal. The value of $Δ E$ is: $(R = 2.02 cal K^{- 1} {mol}^{- 1})$

1

21.2 kcal

2

17.8 kcal

3

18.4 kcal

4

20.6 kcal

Explanation:

We know that, $Δ H = Δ E + Δ n_{g} RT$
$\begin{aligned} Δ n_{g} = 5 - 3 = 2 \\ Δ H = 19 kcal \\ = 19 \times 10^{3} cal . \\ Δ E = Δ H - Δ n_{g} RT \\ = 19 \times 10^{3} - (2 \times 2 \times 300) \\ = 19000 - 1200 = 17800 cal \\ ⊔ 17.8 kcal \end{aligned}$

UPTU/UPSEE-2006

Thermodynamics

273037 At $298 K$ temperature the activation energy for the reaction $x_{2} + y_{2} \to 2 x y + 20 k J$ is $15 kJ$ .
What will be the activation energy for the reaction $2 x y \to x_{2} + y_{2}$ ?

1

- 15 kJ

2

+ 35 kJ

3

- 5 kJ

4

- 35 kJ

Explanation:

$Δ H_{R} =$ activation energy of forward reactionactivation energy of backward reaction-
$\begin{array}{ll} ∵ & - 20 = 15 - x \\ So, & x = + 35 kJ \end{array}$

GUJCET-2017

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