QBManager

CHEMISTRY(KCET)

285281 Conversion of oxygen into ozone is non-spontaneous at

1 all temperatures

2 high temperature

3 room temperature

4 low temperature.

Explanation:

(a) $3 O_{2} ⟶ 2 O_{3}$
$Δ H = + ve : Δ S = - ve$ Equation,
$Δ G = Δ H - T Δ S = + ve - (- ve)$
$Δ G = + ve$ ; Non-spontaneous at all temperatures.

Karnataka CET 2014

CHEMISTRY(KCET)

285276 A reaction has both $Δ H$ and $Δ S$ -ve. The rate of reaction

1 cannot be predicted for change in temperature

2 increases with increase in temperature

3 increases with decrease in temperature

4 remains unaffected by change in temperature.

Explanation:

(c) $As Δ H = - ve$ , so the reaction is exothermic and exothermic reactions are favoured at low temperature.

Karnataka CET 2017

CHEMISTRY(KCET)

285277 If the bond energies of $H - H, Br - Br$ and $H - Br$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, then $Δ H^{\circ}$ 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:

(d) $Δ_{r} H = Σ B . E_{(Reactants)} - Σ B \cdot E_{.(Products)}$
$= [B \cdot E_{(H_{2})} + B \cdot E_{({Br}_{2})}] - [2 B . E_{(HBr)}]$
$= [433 + 192 - 2 \times 364] kJ = - 103 kJ$

Karnataka CET 2016

CHEMISTRY(KCET)

285278 The correct statement regarding entropy is

1 at absolute zero temperature, entropy of a perfectly crystalline solid is zero

2 at absolute zero temperature, the entropy of a perfectly crystalline substance is +ve

3 at absolute zero temperature, the entropy of all crystalline substances is zero

4 at

0^{\circ} C

, the entropy of a perfect crystalline solid is zero.

Karnataka CET 2016

CHEMISTRY(KCET)

285269 Identify the wrong relation for real gases

1

Z = \frac{V_{ideal}}{V_{real}}

2

P_{ideal} = P_{real} + \frac{{an}^{2}}{V^{2}}

3

V_{real} = V_{ideal} - nb

4

(p + \frac{a}{V^{2}}) (V - b) = RT

Explanation:

(a) Compressibility Factor $(Z) = \frac{V_{real}}{V_{ideal}}$

Karnataka CET 2024

CHEMISTRY(KCET)

285281 Conversion of oxygen into ozone is non-spontaneous at

1 all temperatures

2 high temperature

3 room temperature

4 low temperature.

Explanation:

(a) $3 O_{2} ⟶ 2 O_{3}$
$Δ H = + ve : Δ S = - ve$ Equation,
$Δ G = Δ H - T Δ S = + ve - (- ve)$
$Δ G = + ve$ ; Non-spontaneous at all temperatures.

Karnataka CET 2014

CHEMISTRY(KCET)

285276 A reaction has both $Δ H$ and $Δ S$ -ve. The rate of reaction

1 cannot be predicted for change in temperature

2 increases with increase in temperature

3 increases with decrease in temperature

4 remains unaffected by change in temperature.

Explanation:

(c) $As Δ H = - ve$ , so the reaction is exothermic and exothermic reactions are favoured at low temperature.

Karnataka CET 2017

CHEMISTRY(KCET)

285277 If the bond energies of $H - H, Br - Br$ and $H - Br$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, then $Δ H^{\circ}$ 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:

(d) $Δ_{r} H = Σ B . E_{(Reactants)} - Σ B \cdot E_{.(Products)}$
$= [B \cdot E_{(H_{2})} + B \cdot E_{({Br}_{2})}] - [2 B . E_{(HBr)}]$
$= [433 + 192 - 2 \times 364] kJ = - 103 kJ$

Karnataka CET 2016

CHEMISTRY(KCET)

285278 The correct statement regarding entropy is

1 at absolute zero temperature, entropy of a perfectly crystalline solid is zero

2 at absolute zero temperature, the entropy of a perfectly crystalline substance is +ve

3 at absolute zero temperature, the entropy of all crystalline substances is zero

4 at

0^{\circ} C

, the entropy of a perfect crystalline solid is zero.

Karnataka CET 2016

CHEMISTRY(KCET)

285269 Identify the wrong relation for real gases

1

Z = \frac{V_{ideal}}{V_{real}}

2

P_{ideal} = P_{real} + \frac{{an}^{2}}{V^{2}}

3

V_{real} = V_{ideal} - nb

4

(p + \frac{a}{V^{2}}) (V - b) = RT

Explanation:

(a) Compressibility Factor $(Z) = \frac{V_{real}}{V_{ideal}}$

Karnataka CET 2024

CHEMISTRY(KCET)

285281 Conversion of oxygen into ozone is non-spontaneous at

1 all temperatures

2 high temperature

3 room temperature

4 low temperature.

Explanation:

(a) $3 O_{2} ⟶ 2 O_{3}$
$Δ H = + ve : Δ S = - ve$ Equation,
$Δ G = Δ H - T Δ S = + ve - (- ve)$
$Δ G = + ve$ ; Non-spontaneous at all temperatures.

Karnataka CET 2014

CHEMISTRY(KCET)

285276 A reaction has both $Δ H$ and $Δ S$ -ve. The rate of reaction

1 cannot be predicted for change in temperature

2 increases with increase in temperature

3 increases with decrease in temperature

4 remains unaffected by change in temperature.

Explanation:

(c) $As Δ H = - ve$ , so the reaction is exothermic and exothermic reactions are favoured at low temperature.

Karnataka CET 2017

CHEMISTRY(KCET)

285277 If the bond energies of $H - H, Br - Br$ and $H - Br$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, then $Δ H^{\circ}$ 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:

(d) $Δ_{r} H = Σ B . E_{(Reactants)} - Σ B \cdot E_{.(Products)}$
$= [B \cdot E_{(H_{2})} + B \cdot E_{({Br}_{2})}] - [2 B . E_{(HBr)}]$
$= [433 + 192 - 2 \times 364] kJ = - 103 kJ$

Karnataka CET 2016

CHEMISTRY(KCET)

285278 The correct statement regarding entropy is

1 at absolute zero temperature, entropy of a perfectly crystalline solid is zero

2 at absolute zero temperature, the entropy of a perfectly crystalline substance is +ve

3 at absolute zero temperature, the entropy of all crystalline substances is zero

4 at

0^{\circ} C

, the entropy of a perfect crystalline solid is zero.

Karnataka CET 2016

CHEMISTRY(KCET)

285269 Identify the wrong relation for real gases

1

Z = \frac{V_{ideal}}{V_{real}}

2

P_{ideal} = P_{real} + \frac{{an}^{2}}{V^{2}}

3

V_{real} = V_{ideal} - nb

4

(p + \frac{a}{V^{2}}) (V - b) = RT

Explanation:

(a) Compressibility Factor $(Z) = \frac{V_{real}}{V_{ideal}}$

Karnataka CET 2024

CHEMISTRY(KCET)

285281 Conversion of oxygen into ozone is non-spontaneous at

1 all temperatures

2 high temperature

3 room temperature

4 low temperature.

Explanation:

(a) $3 O_{2} ⟶ 2 O_{3}$
$Δ H = + ve : Δ S = - ve$ Equation,
$Δ G = Δ H - T Δ S = + ve - (- ve)$
$Δ G = + ve$ ; Non-spontaneous at all temperatures.

Karnataka CET 2014

CHEMISTRY(KCET)

285276 A reaction has both $Δ H$ and $Δ S$ -ve. The rate of reaction

1 cannot be predicted for change in temperature

2 increases with increase in temperature

3 increases with decrease in temperature

4 remains unaffected by change in temperature.

Explanation:

(c) $As Δ H = - ve$ , so the reaction is exothermic and exothermic reactions are favoured at low temperature.

Karnataka CET 2017

CHEMISTRY(KCET)

285277 If the bond energies of $H - H, Br - Br$ and $H - Br$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, then $Δ H^{\circ}$ 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:

(d) $Δ_{r} H = Σ B . E_{(Reactants)} - Σ B \cdot E_{.(Products)}$
$= [B \cdot E_{(H_{2})} + B \cdot E_{({Br}_{2})}] - [2 B . E_{(HBr)}]$
$= [433 + 192 - 2 \times 364] kJ = - 103 kJ$

Karnataka CET 2016

CHEMISTRY(KCET)

285278 The correct statement regarding entropy is

1 at absolute zero temperature, entropy of a perfectly crystalline solid is zero

2 at absolute zero temperature, the entropy of a perfectly crystalline substance is +ve

3 at absolute zero temperature, the entropy of all crystalline substances is zero

4 at

0^{\circ} C

, the entropy of a perfect crystalline solid is zero.

Karnataka CET 2016

CHEMISTRY(KCET)

285269 Identify the wrong relation for real gases

1

Z = \frac{V_{ideal}}{V_{real}}

2

P_{ideal} = P_{real} + \frac{{an}^{2}}{V^{2}}

3

V_{real} = V_{ideal} - nb

4

(p + \frac{a}{V^{2}}) (V - b) = RT

Explanation:

(a) Compressibility Factor $(Z) = \frac{V_{real}}{V_{ideal}}$

Karnataka CET 2024

CHEMISTRY(KCET)

285281 Conversion of oxygen into ozone is non-spontaneous at

1 all temperatures

2 high temperature

3 room temperature

4 low temperature.

Explanation:

(a) $3 O_{2} ⟶ 2 O_{3}$
$Δ H = + ve : Δ S = - ve$ Equation,
$Δ G = Δ H - T Δ S = + ve - (- ve)$
$Δ G = + ve$ ; Non-spontaneous at all temperatures.

Karnataka CET 2014

CHEMISTRY(KCET)

285276 A reaction has both $Δ H$ and $Δ S$ -ve. The rate of reaction

1 cannot be predicted for change in temperature

2 increases with increase in temperature

3 increases with decrease in temperature

4 remains unaffected by change in temperature.

Explanation:

(c) $As Δ H = - ve$ , so the reaction is exothermic and exothermic reactions are favoured at low temperature.

Karnataka CET 2017

CHEMISTRY(KCET)

285277 If the bond energies of $H - H, Br - Br$ and $H - Br$ are 433,192 and $364 kJ {mol}^{- 1}$ respectively, then $Δ H^{\circ}$ 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:

(d) $Δ_{r} H = Σ B . E_{(Reactants)} - Σ B \cdot E_{.(Products)}$
$= [B \cdot E_{(H_{2})} + B \cdot E_{({Br}_{2})}] - [2 B . E_{(HBr)}]$
$= [433 + 192 - 2 \times 364] kJ = - 103 kJ$

Karnataka CET 2016

CHEMISTRY(KCET)

285278 The correct statement regarding entropy is

1 at absolute zero temperature, entropy of a perfectly crystalline solid is zero

2 at absolute zero temperature, the entropy of a perfectly crystalline substance is +ve

3 at absolute zero temperature, the entropy of all crystalline substances is zero

4 at

0^{\circ} C

, the entropy of a perfect crystalline solid is zero.

Karnataka CET 2016

CHEMISTRY(KCET)

285269 Identify the wrong relation for real gases

1

Z = \frac{V_{ideal}}{V_{real}}

2

P_{ideal} = P_{real} + \frac{{an}^{2}}{V^{2}}

3

V_{real} = V_{ideal} - nb

4

(p + \frac{a}{V^{2}}) (V - b) = RT

Explanation:

(a) Compressibility Factor $(Z) = \frac{V_{real}}{V_{ideal}}$

Karnataka CET 2024

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